Though it seems long

Similar solution as Vighnesh Shenoy but different presentation:

$\begin{aligned} \sum_{x=1}^\infty \prod_{n=0}^9 \frac{1}{x+n} & = \sum_{x=1}^\infty \frac{(x-1)!}{(x+9)!} \quad \quad \color{#3D99F6}{\text{See Note }} \\ & = \frac{1}{9} \sum_{x=1}^\infty \left(\frac{(x-1)!}{(x+8)!}-\frac{x!}{(x+9)!}\right) \\ & = \frac{1}{9}\left( \sum_{x=1}^\infty \frac{(x-1)!}{(x+8)!}- \sum_{x=1}^\infty \frac{x!}{(x+9)!}\right) \\ & = \frac{1}{9}\left( \sum_{x=1}^\infty \frac{(x-1)!}{(x+8)!}- \sum_{x=2}^\infty \frac{(x-1)!}{(x+8)!}\right) \\ & = \frac{1}{9}\left(\frac{0!}{9!}\right) \\ & = \frac{1}{9 \times 9!} \end{aligned}$

$\Rightarrow A = 9 \times 9! = \boxed{3265920}$

$\color{#3D99F6}{\text{Note:}}$

$\begin{aligned} \frac{(x-1)!}{(x+8)!}-\frac{x!}{(x+9)!} & = \frac{(x-1)!(x+9)-x!}{(x+9)!} = \frac{x! + 9(x-1)!-x!}{(x+9)!} = \frac{9(x-1)!}{(x+9)!} \end{aligned}$

$\displaystyle S = \sum_{x=1}^{\infty}\prod_{n=0}^{9} \dfrac{1}{x+n}$

$S < \sum_{n=1}^{\infty} \dfrac{1}{(n)(n+1)} = 1$
Thus the series is absolutely convergent and any rearrangement is justfied.

$\displaystyle \therefore S = \sum_{x=1}^{\infty} \dfrac{1}{x \times (x+1) \times (x+2) ... \times (x+9) }$
$\displaystyle \therefore S = \sum_{x=1}^{\infty} \dfrac{1}{x \times (x+1) \times (x+2) ... \times (x+9) }$
$\displaystyle \therefore S = \dfrac{1}{9} \left( \times \sum_{x=1}^{\infty} \dfrac{1}{x \times (x+1)...\times(x+8)} - \dfrac{1}{(x+1)\times(x+2)...\times(x+9)} \right)$

The given summation forms a telescopic series where all the terms cancel out except for the first and the last.
$\displaystyle \therefore S = \dfrac{1}{9}\times \left( \dfrac{1}{9!} -\lim_{x->\infty} \dfrac{1}{(x+1)\times (x+2)...\times (x+9)} \right) = \dfrac{1}{9 \times 9!} - 0$

$\displaystyle \therefore A = 9 \times 9! =3265920$