It's a Trap!

$\begin{aligned} P & = \prod_{n=1}^\infty \frac {\lfloor n! {\color{#3D99F6} e} \rfloor}{\lfloor n!{\color{#3D99F6} e}-1\rfloor} & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \prod_{n=1}^\infty \frac {\left \lfloor n! {\color{#3D99F6} \sum_{k=0}^\infty \frac 1{k!}} \right \rfloor}{\left \lfloor n!{\color{#3D99F6} \sum_{k=0}^\infty \frac 1{k!}}-1 \right \rfloor} \\ & = \prod_{n=1}^\infty \frac {\sum_{k=0}^n \frac {n!}{k!}}{\sum_{k=0}^n \frac {n!}{k!}-1} \\ & = \prod_{n=1}^\infty \left(1+\frac 1{\sum_{k=0}^n \frac {n!}{k!}-1} \right) \\ & = \prod_{n=1}^\infty \left(1+\frac 1{\color{#3D99F6}\sum_{k=0}^{n-1} \frac {n!}{k!}} \right) & \small \color{#3D99F6} \text{Since }a_n = \sum_{k=0}^{n-1} \frac {n!}{k!} \text{ satisfies} \\ & = \prod_{n=1}^\infty \left(1+\frac 1{\color{#3D99F6}a_n} \right) & \small \color{#3D99F6} a_n = n(a_{n-1}+1) \\ & = e & \small \color{#3D99F6} \text{See note below,} \end{aligned}$

Therefore $\ln P = \boxed 1$ .

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Note: If $a_n$ satisfies the recurrent relation $a_n = n(a_{n-1}+1)$ and $a_1 = a^{-1}$ , then $\displaystyle \prod_{n=1}^\infty (1+a_n^{-1}) = e$ . See reference (equations (24) and (25)) . Now let us check if $\displaystyle a_n = \sum_{k=0}^{n-1} \frac {n!}{k!}$ satisfies the recurrent relation.

For $n=1$ , $\displaystyle a_1 = \sum_{k=0}^{1-1} \frac {n!}{k!} = \frac {1!}{0!} = 1$ and $n=2$ , $\displaystyle a_2 = \frac {2!}{0!} + \frac {2!}{1!} = 4 = 2(a_1 + 1)$ so the claim is true for $n=2$ . Assuming the claim is true for $n$ , then $\displaystyle a_{n+1} = \sum_{k=0}^n \frac {(n+1)!}{k!} = (n+1) \sum_{k=0}^{n-1} \frac {n!}{k!} + (n+1)\frac {n!}{n!} = (n+1)(a_n + 1)$ . Therefore, the claim is also true for $n+1$ and hence true for all $n \ge 1$ .

I started out like Cde Cheong but took a different turn after his third line.

$\Large P_N =\prod_{n=1}^N\frac{\sum_{k=0}^n\frac{n!}{k!}}{\sum_{k=0}^n\frac{n!}{k!}-1}= \prod_{n=1}^N\frac{\sum_{k=0}^n\frac{n!}{k!}}{\sum_{k=0}^{n-1}\frac{n!}{k!}}=\prod_{n=1}^N\frac{\sum_{k=0}^n\frac{1}{k!}}{\sum_{k=0}^{n-1}\frac{1}{k!}}= \sum_{k=0}^{N}\frac{1}{k!}.$ In the last equation, we make use of a telescoping product.

Now $P=\lim_{N\to\infty}P_N = e$ and $\ln(P)=\boxed1$