Let $s_n=\sum\limits_{k=0}^n\frac{1}{k!}$ and $b_n=\left(1+\frac{1}{n}\right)^n$. You have shown that
\begin{align}
s_n\geq b_n.\tag{@}
\end{align}
You can easily prove that $s_n$ is increasing with $n$ (so the limit exists in $[0,\infty]$). By the binomial expansion, you also showed that
\begin{align}
b_n=\sum\limits_{k=0}^n\frac{1}{k!}\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right).\tag{$*$}
\end{align}
The equality $(*)$ has two consequences.
First, we have $b_n$ is increasing because
\begin{align}
b_{n+1}&= \sum\limits_{k=0}^{n+1}\frac{1}{k!}\prod_{j=0}^{k-1}\left(1-\frac{j}{n+1}\right)\\
&\geq \sum\limits_{k=0}^{n+1}\frac{1}{k!}\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right)\\
&\geq \sum\limits_{k=0}^n\frac{1}{k!}\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right)\\
&=b_n
\end{align}
Hence, the limit of $b_n$ exists (a-priori, in $[0,\infty]$) because it equals its supremum.
Next, we have that for any $n\geq m$,
\begin{align}
b_n&\geq \sum\limits_{k=0}^m\frac{1}{k!}\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right).
\end{align}
By the first bullet point the limit as $n\to\infty$ of $b_n$ exists, while on the RHS, the limit as $n\to\infty$ also clearly exists and equals $s_m$:
\begin{align}
\lim_{n\to\infty}b_n&\geq s_m.\tag{@@}
\end{align}
Finally, combining (@) with (@@) gives
\begin{align}
\lim_{n\to\infty}b_n\geq \lim_{m\to\infty}s_m\geq \lim_{m\to\infty}b_m.
\end{align}
The first and third terms are equal, so we have equalities everywhere:
\begin{align}
\lim_{n\to\infty}b_n=\lim_{n\to\infty}s_n.
\end{align}
Finally, for the sake of completeness, we can observe using the series definition that
\begin{align}
2<e=\sum_{k=0}^{\infty}\frac{1}{k!}=1+1+\sum_{k=2}^{\infty}\frac{1}{k!}<1+1+\sum_{k=2}^{\infty}\frac{1}{2^{k-1}}=1+1+1=3,
\end{align}
where we have summed the geometric series (which is of course the key fact underlying the ratio/root tests and hence why the series definition of $e$, or more generally the series definition of the exponential function converges absolutely and uniformly on compact subsets of the complex plane); and thus we see that $2< e<3$; in particular, it is finite.
This is a somewhat common question (I see 15 related links), as you can see from the links below, but I still answered it because I think this is a direct and elementary enough proof (using only that non-negative increasing sequences have limits in $[0,\infty]$) which no one seems to have written before. NOTE: I am obviously not claiming originality here, I’m just saying no one bothers to write it; instead they opt to prove more in the process by defining corresponding functions for all $z\in\Bbb{C}$, then finally plug in $z=1$)
- Prove the definitions of $e$ to be equivalent. The title is misleading; it’s more like showing two different expressions agree for all $x\in\Bbb{R}$. A-posteriori, one can deduce the equality when $x=1$.
- Is there a direct way to prove $\lim\limits_{n\to\infty}(1+\frac{1}{n})^n=\sum\limits_{n=0}\frac{1}{n!}$? This was closed as a dupe (but I don’t see why). The answer by @SomeStrangeUser reproduces what is in Rudin, and is of course a nice proof, but here I wanted to avoid stuff with $\limsup$ and $\liminf$, and I wanted to give a self-contained explanation for why all the limits exist (something which everyone takes for granted… probably because this isn’t a new result, but still, I think it’s worth writing down explicitly and directly).
- Prove that $\lim_{n \to\infty}(1+\frac{1}{n})^n =\lim_{n \to\infty} (\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots + \frac{1}{n!})$ Also closed as a duplicate linking to the first link. But, I should say that the answer given there is a nice and direct (outline), exploiting both increasing and decreasing sequences (but for fun, I wanted to write a proof using only increasing sequences).
- Why does the sum of the reciprocals of factorials converge to $e$? This is one of the oldest and most viewed, but the accepted answer is very incomplete, and the others exploit fanicer techniques with series etc, so again not really direct.
So, the only merit of the answer I gave here is ‘directness’, but I should say that the above proof is not even my favourite. By the wealth of other answers seen in the various links (and sublinks) above, there are multiple proofs of this fact (some very slick ones involving things like DCT), and often establish more in the process, so there’s a lot to be learned from there as well.
