<s>[INST] Proving the convergence of an infinite product I’m trying to prove the Taylor series of $e$ using binomial expansion:

$$e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$$

The steps I’ve tried so far are:
$$\left(1+\frac{1}{n}\right)^n \\
= 1 + \binom{n}{1}\frac{1}{n} + \binom{n}{2}\frac{1}{n^2} + \binom{n}{3}\frac{1}{n^3} + \dots + \binom{n}{n}\frac{1}{n^n} \\
= 1+\dfrac {n}{1!}\left( \dfrac {1}{n}\right) +\dfrac {n\left( n-1\right) }{2!}\left( \dfrac {1}{n^{2}}\right) + \dfrac {n\left( n-1\right) \left( n-2\right) }{3!}\left( \dfrac {1}{n^{3}}\right) +\dots +\dfrac {n\left( n-1\right) \left( n-2\right) \dots 1}{n!}\left( \dfrac {1}{n^{n}}\right) \\
= 1+\dfrac {1}{1!}+\dfrac {1}{2!}\left( 1-\dfrac {1}{n}\right) +\dfrac {1}{3!}\left( 1-\dfrac {1}{n}\right) \left( 1-\dfrac {2}{n}\right) +\ldots +\dfrac {1}{n!}\left( 1-\dfrac {1}{n}\right) \left( 1-\dfrac {2}{n}\right) \dots
\left( 1-\dfrac {n-1}{n}\right) $$
As $n \to \infty$, the last term becomes 
$$\lim_{n\to\infty} \left(\frac{1}{n!} \prod_{k=1}^n \left(1-\frac{k-1}{n}\right)\right)$$ and the infinite product on the right of the term should be proven to be convergent and equals to $1$. So how do I prove that?
 [/INST] $$\prod_{k=1}^n\left(1-\frac{k-1}{n}\right)=\prod_{k=1}^n\frac{1}{n}\left(n-k+1\right)=\frac{1}{n^n}\big[n(n-1)\cdots(1)\big]=\frac{n!}{n^n}$$
Hence, we have:
$$\lim_{n\to\infty}\frac{1}{n!}\prod_{k=1}^n\left(1-\frac{k-1}{n}\right)=\lim_{n\to\infty}\frac{n!}{n^n\cdot n!}=\lim_{n\to\infty}\frac{1}{n^n}=0$$
This should be expected, as if the last term approached any other value then the terms of the summation would not approach zero.
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