How to derive $\lim\limits_{x,n→∞}\prod\limits_{k=0}^{n-1}\left(1-\frac kx\right)=1-\frac eα\frac1x$ where $α=2e-4$? Background
$\def\d{\mathrm{d}}\def\e{\mathrm{e}}\def\i{\mathrm{i}}$
The following two equations are well known,
$$\begin{align*}
\left(1+\frac1x\right)^x &= \sum_{n=0}^\infty {x \choose n}\frac1{x^n}, \\[1ex]
\e &= \sum_{n=0}^\infty\dfrac{1}{n!}.
\end{align*}$$
Subtract them, which will produce the product in the question, and then perform a backward derivation (see end for details) from
$$\lim_{x\to\infty}x\left[\left( 1+\dfrac{1}{x} \right)^{x}-\e \right] = -\dfrac{\e}{2},$$
yields the result in the question.

P.S. The limit above comes from a small exercise, its solution steps are:
   ① Reciprocal substitution, i.e. $x=\dfrac1t$;
   ② Use L'Hôpital's rule once;
   ③ Utilize the derivation formula for compound logarithmic functions, i.e. $(u^v)' = u^v \left( \ln u^{v'} + \dfrac{u'}{u}v \right)$;
   ④ Use L'Hôpital's rule twice.


Q
How to solve
$$\lim_{\substack{n\to\infty\\x\to\infty}}\prod_{k=0}^{n-1}\left(1-\frac kx\right)$$
in a forward approach? (while requiring the retention of the "exact form" in the title)
Actually, what I'd like to ask more is: Is there a general solution for products of this form? In other words, is it possible to assume that the result is in the form containing {$\e, \alpha$} and find it?



Side note
The specific process of backward derivation
$$\begin{align*}
\Sigma_1-\Sigma_2&=\sum_{n=0}^1\,(1-1) + \sum_{n=2}^{\infty}\left( \frac{x(x-1)\cdots(x-(n-1))}{n!}\frac{1}{x^n}-\frac{1}{n!} \right) \\
&=\sum_{n=2}^{\infty}\dfrac{1}{n!}\left[ 1\left( 1-\dfrac{1}{x} \right)\cdots\left( 1-\dfrac{n-1}{x} \right) - 1 \right] \\
&=\sum_{n=2}^{\infty}\dfrac{1}{n!}\left[ {\color{teal}{\prod_{k=0}^{n-1}\left( 1-\dfrac{k}{x} \right) - 1}} \right],
\end{align*}$$
Substitute into the equation in the question, immediately get
$$\lim_{x\to\infty} x\,\left(\Sigma_1-\Sigma_2\right) = {\color{teal}{-\dfrac{\e}{2}\dfrac{1}{\e-2}}}\sum_{n=2}^{\infty}\dfrac{1}{n!}=-\dfrac{\e}{2}.$$
The result is consistent with the known conclusions from the normal calculation, so the equation in question is verified.

The product $$ \prod_{k=0}^{n-1} \left( 1 - \frac{k}{x} \right) $$
can be seen as
$$ \prod_{k=0}^{n-1} \left( 1 - \frac{k}{x} \right) = \frac{x (x-1) (x-2)\cdots (x-n+1)}{x^n} = \frac{x!}{x^n \, (x-n)!}. $$
Using $n! \approx \sqrt{2 \pi n} \, n^n \, e^{-n}$ then
$$ \prod_{k=0}^{n-1} \left( 1 - \frac{k}{x} \right) = \frac{x!}{x^n \, (x-n)!} \approx \left( 1- \frac{n}{x}\right)^{n-x-1/2} \, e^{-n}. $$
If $ n \to \infty$ then
$$ \lim_{n \to \infty} \, \prod_{k=0}^{n-1} \left( 1 - \frac{k}{x} \right) =  0. $$
If $x \to \infty$ then, by use of $\lim_{x \to \infty} \left(1 - \frac{n}{x}\right)^{-x} = e^{n}$,
$$ \lim_{x \to \infty} \, \prod_{k=0}^{n-1} \left( 1 - \frac{k}{x} \right) = e^{-n} \, \lim_{x \to \infty} \left( 1 - \frac{n}{x}\right)^{n-1/2} \, \left(1 - \frac{n}{x}\right)^{-x} = 1. $$
Additional note:
Consider the limit
$$ \lim_{x \to \infty} \, x \, \left[ \left(1 - \frac{a}{x}\right)^x - e^{-a} \right]$$
as follows.
\begin{align}
\left(1 - \frac{a}{x}\right)^x - e^{-a} &= e^{x \, \ln\left(1 - \frac{a}{x}\right)} - e^{-a} \\
&= e^{x \, \left( - \frac{a}{x} \right) - \frac{a^2}{2 \, x^2} + \cdots} - e^{-a} \\
&= e^{-a - \frac{a^2}{2 \, x} - \cdots} - e^{-a} = e^{-a} \, \left( -1 + e^{\frac{a^2}{2 \, x} + \cdots} \right) \\
&= e^{-a} \, \left( -1 + 1 + \left( -\frac{a^2}{2 \, x} - \frac{a^3}{3! \, x^2} + \cdots \right) + \frac{1}{2!} \, \left( -\frac{a^2}{2 \, x} - \frac{a^3}{3! \, x^2} + \cdots \right)^2 + \cdots \right) \\
&= e^{-a} \, \left( -\frac{a^2}{2 \, x} + \frac{a^3 \, (3 a - 4)}{4! \, x^2} + \mathcal{O}\left(\frac{1}{x^3}\right) \right) \\
&= -\frac{a^2 \, e^{-a}}{2 \, x} \, \left( 1 + \frac{2 a (4 - 3 a)}{4! \, x} + \mathcal{O}\left(\frac{1}{x^2}\right) \right).
\end{align}
This leads to
\begin{align}
\lim_{x \to \infty} \, x \, \left[\left(1 - \frac{a}{x}\right)^x - e^{-a} \right] &= \lim_{x \to \infty} \, -\frac{a^2 \, e^{-a}}{2} \, \left( 1 + \frac{2 a (4 - 3 a)}{4! \, x} + \mathcal{O}\left(\frac{1}{x^2}\right) \right) \\
&= -\frac{a^2 \, e^{-a}}{2}.
\end{align}
