I Was Very Amazed At The Solution 8

This solution requires nothing but a subtle observation. We let $\begin{aligned} S&=&\sum^{\infty}_{n=1}\frac{(n-1)!}{\prod^n_{i=1}(9+i)}\\ &=&{\color{#3D99F6}\sum^{\infty}_{n=1}\frac{(n-1)!\times9!}{(n+9)!}}\\ &=&\frac{1}{10}\sum^{\infty}_{n=1}\frac{(n-1)!\times9!\times10}{(n+9)!}\\ &=&\frac{1}{10}\sum^{\infty}_{n=1}\frac{(n-1)!\times9!\times(n+9-(n-1))}{(n+9)!}\\ &=&\frac{1}{10}\sum^{\infty}_{n=1}\left(\frac{(n-1)!\times9!\times(n+9)}{(n+9)!}-\frac{(n-1)!\times(n-1)\times9!}{(n+9)!}\right)\\ &=&\frac{1}{10}\sum^{\infty}_{n=1}\left(\frac{(n-1)!\times9!}{(n+8)!}-\frac{n!\times9!}{(n+9)!}+\frac{(n-1)!\times9!}{(n+9)!}\right)\\ &=&\frac{1}{10}\sum^{\infty}_{n=1}\left({\color{magenta}\frac{(n-1)!\times9!}{(n+8)!}-\frac{n!\times9!}{(n+9)!}}\right)+{\color{#3D99F6}\frac{1}{10}\sum^{\infty}_{n=1}\frac{(n-1)!\times9!}{(n+9)!}} \end{aligned}$ Observe that at the last step, we used the equality $(n-1)\times(n-1)!=n!-(n-1)!$ . Also, notice that the terms highlighted in blue are multiples of each other, and the terms highlighted in magenta is telescoping. With that, we obtain $9S=1,$ which gives $81S=\boxed{9}.$

There is a simpler answer with no need of beta function:

$\large \displaystyle S = 81 \sum_{n=1}^{\infty} \frac{(n-1)!}{\prod_{i=1}^n (9+i)}$

$\large \displaystyle S = 81 \cdot \color{#3D99F6} 9! \color{#333333} \sum_{n=1}^{\infty} \frac{(n-1)!}{\color{#3D99F6} 9! \color{#333333}\prod_{i=1}^n (9+i)}$

$\large \displaystyle S = 81 \cdot 9! \sum_{n=1}^{\infty} \frac{(n-1)!}{(n+9)!}$

$\large \displaystyle S = 81 \cdot 9! \sum_{n=1}^{\infty} \frac{1}{(n+9)(n+8)(n+7)(n+6)(n+5)(n+4)(n+3)(n+2)(n+1)n}$

$\large \displaystyle S = 81 \cdot 9! \sum_{n=1}^{\infty} \frac{1}{(n+9)(n+8)(n+7)(n+6)(n+5)(n+4)(n+3)(n+2)(n+1)n} \color{#3D99F6} \cdot \frac{(n+9) - n}{9}$

$\large \displaystyle S = 9\cdot 9! \sum_{n=1}^{\infty} \frac{1}{(n+8)(n+7)(n+6)(n+5)(n+4)(n+3)(n+2)(n+1)n} - \frac{1}{(n+9)(n+8)(n+7)(n+6)(n+5)(n+4)(n+3)(n+2)(n+1)}$

This will telescope, and only the first term will remain:

$\large \displaystyle S = 9\cdot 9! \left [ \frac{1}{(n+8)(n+7)(n+6)(n+5)(n+4)(n+3)(n+2)(n+1)n} \right]_{n=1}$

$\large \displaystyle S = 9\cdot 9! \cdot \frac{1}{9!}$

$\color{#3D99F6} \boxed{\large \displaystyle S = 9}$

We have $\int_0^1 x^{a-1}(1-x)^{b-1}\,dx \; = \; B(a,b) \; = \; \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} \; = \; \frac{(a-1)! (b-1)!}{(a+b-1)!}$ for integers $a,b \ge 1$ . We need to evaluate $\begin{aligned} 81\sum_{n=1}^\infty \frac{(n-1)! 9!}{(n+9)!} & = \; 81\sum_{n=1}^\infty B(n,10) \; = \; 81\sum_{n=1}^\infty \int_0^1 x^{n-1}(1-x)^9\,dx \\ & = \; 81 \int_0^1 \frac{1}{1-x} (1-x)^9\,dx \; = \; 81\int_0^1 (1-x)^8\,dx \; = \; 81 \times \tfrac19 \; = \; \boxed{9} \end{aligned}$

I try it for certain values of $\alpha$ i.e. 1, 2, 5, 9 and my theorem somewhat works.

$\sum_{n = 1}^{\infty} \dfrac{(n-1)!}{\displaystyle \prod_{i = 1}^{n} ( \alpha + i )} = \dfrac{1}{\alpha}$

and hence,

$\displaystyle 81\sum_{n = 1}^{\infty} \dfrac{(n-1)!}{\displaystyle \prod_{i = 1}^{n} ( 9 + i )} = 81\cdot\dfrac{1}{9} = \boxed{9}$

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For $\alpha = 1$

$\begin{aligned} \displaystyle \sum_{n = 1}^{\infty} \dfrac{(n-1)!}{\displaystyle \prod_{i = 1}^{n} ( 1 + i )} &= 1!\sum_{n=0}^{\infty} \dfrac{n!}{(n+2)!} \\ &= \sum_{n=0}^{\infty} \dfrac{1!}{(n+2)(n+1)} \\ &= \sum_{n=0}^{\infty} \left( \dfrac{\color{#D61F06}{1}}{n+1} - \dfrac{\color{#D61F06}{1}}{(n+2)} \right) \\ &= \dfrac{1}{1} \end{aligned}$

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For $\alpha = 2$

$\begin{aligned} \displaystyle \sum_{n = 1}^{\infty} \dfrac{(n-1)!}{\displaystyle \prod_{i = 1}^{n} ( 2 + i )} &= 2!\sum_{n=0}^{\infty} \dfrac{n!}{(n+3)!} \\ &= \sum_{n=0}^{\infty} \dfrac{2!}{(n+3)(n+2)(n+1)} \\ &= \sum_{n=0}^{\infty} \left( \dfrac{\color{#D61F06}{1}}{n+1} - \dfrac{\color{#D61F06}{2}}{(n+2)} + \dfrac{\color{#D61F06}{1}}{(n+3)}\right) \\ &= \dfrac{1}{2} \end{aligned}$

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For $\alpha = 5$

$\begin{aligned} \displaystyle \sum_{n = 1}^{\infty} \dfrac{(n-1)!}{\displaystyle \prod_{i = 1}^{n} ( 5 + i )} &= 5!\sum_{n=0}^{\infty} \dfrac{n!}{(n+6)!} \\ &= \sum_{n=0}^{\infty} \dfrac{5!}{(n+6)(n+5)(n+4)(n+3)(n+2)(n+1)} \\ &= \sum_{n=0}^{\infty} \left( \dfrac{\color{#D61F06}{1}}{n+1} - \dfrac{\color{#D61F06}{5}}{(n+2)} + \dfrac{\color{#D61F06}{10}}{(n+3)} - \dfrac{\color{#D61F06}{10}}{n+4} - \dfrac{\color{#D61F06}{5}}{(n+5)} + \dfrac{\color{#D61F06}{10}}{(n+6)}\right) \\ &= \dfrac{1}{5} \end{aligned}$

and so on...

Notice that the red colored numbers are the numbers in the $(\alpha + 1)th$ row of the Pascal's Triangle.

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Update: Formal Proof of my Theorem

$\begin{aligned} \sum_{n = 1}^{\infty} \dfrac{(n-1)!}{\displaystyle \prod_{i = 1}^{n} ( \alpha + i )} \\ &= \sum_{n = 1}^{\infty} \dfrac{(n-1)! \cdot \alpha}{\displaystyle \alpha (\alpha + 1) (\alpha + 2) \cdots (\alpha + n)} \\ &= \sum_{n = 1}^{\infty} \dfrac{(n-1)! \cdot (\alpha + n - n)}{\displaystyle \alpha (\alpha + 1) (\alpha + 2) \cdots (\alpha + n)} \\ &= \sum_{n = 1}^{\infty} \dfrac{\big((n-1)! \ (\alpha + n)\big) - (n!)}{\displaystyle \alpha (\alpha + 1) (\alpha + 2) \cdots (\alpha + n - 1) (\alpha + n)} \\ &= \sum_{n = 1}^{\infty} \left( \dfrac{(n-1)!}{\displaystyle \alpha (\alpha + 1) (\alpha + 2) \cdots (\alpha + n-1)} - \dfrac{(n)!}{\displaystyle \alpha (\alpha + 1) (\alpha + 2) \cdots (\alpha + n)} \right) \cdot \end{aligned}$

Let $\displaystyle f(n) = \dfrac{(n-1)!}{\displaystyle \alpha (\alpha + 1) (\alpha + 2) \cdots (\alpha + n-1)} \implies f(n+1) = \dfrac{(n)!}{\displaystyle \alpha (\alpha + 1) (\alpha + 2) \cdots (\alpha + n)} \cdot$

$\begin{aligned}\implies \sum_{n = 1}^{\infty} \dfrac{(n-1)!}{\displaystyle \prod_{i = 1}^{n} ( \alpha + i )} \\ &= \sum_{n = 1}^{\infty} (f(n) - f(n+1)) \\ &= \big(f(1) - f(2)\big) + \big(f(2) - f(3)\big) + \big(f(3) - f(4)\big) + \cdots \end{aligned}$

which will telescopes to $f(1) \cdot$

$\implies \sum_{n = 1}^{\infty} \dfrac{(n-1)!}{\displaystyle \prod_{i = 1}^{n} ( \alpha + i )} = f(1) = \dfrac{(1 - 1)!}{\alpha} = \dfrac{1}{\alpha} \cdot$

Credits to the Owner of this Solution. ^^