Prove that $\lim_{n \to\infty}(1+\frac{1}{n})^n =\lim_{n \to\infty} (\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots + \frac{1}{n!})$

Question

I am trying to prove that this equality is correct. \begin{equation} \lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n =\lim_{n \to\infty} \left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots + \frac{1}{n!}\right) \end{equation} I know that left and right hand side is equal to $e$, but as mathematician I would like to see is this two definition of $e$ is equal.
My idea is to try avoid using hard calculus tools, and try two show that two sequence has the same limit.

Answer 1

As a hint you might start by proving that: $$ \left(1+{1\over n}\right)^n \le \sum_{k=0}^n{1\over k!} \le \left(1 + {1\over n }\right)^{n+1} $$

After which the result follows, since LHS is monotonically increasing, RHS is decreasing, both LHS and RHS are bounded hence: $$ e = \lim_{n\to\infty}\left(1+{1\over n}\right)^n \le \lim_{n\to\infty}\sum_{k=0}^n{1\over k!} \le \lim_{n\to\infty}\left(1 + {1\over n }\right)^{n+1} = e $$