an expression for the $ e^x $ using the binomial theorem Is it possible using the Binomial theorem , to prove the identity
$$ e \sim \left(1+\frac{1}{n}\right)^\frac{1}{n}\sim \sum_{k=0}^n\frac{1}{k!} $$
where $ n \to \infty $
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
& \color{#44f}{\pars{1 + {1 \over n}}^{n}} =
\sum_{k = 0}^{n}{n \choose k}
\pars{1 \over n}^{k} =
\sum_{k = 0}^{n}{1 \over k!}
\,{n! \over \pars{n - k}!n^{k}}
\\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\hspace{3mm}
\sum_{k = 0}^{n}{1 \over k!}
\,{\root{2\pi}n^{n + 1/2}\,\,\expo{-n} \over
\bracks{\root{2\pi}
\pars{n - k}^{n - k + 1/2}\,\,\,
\expo{-\pars{n - k}}\,}n^{k}}
\\[5mm] = & \
\sum_{k = 0}^{n}{1 \over k!}\,{\expo{-k} \over
\bracks{n^{-k}\,\,\pars{1 - k/n}^{n}\,\,
\pars{1 - k/n}^{1/2 - k}\,\,}n^{k}}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim} & \hspace{5mm}
\sum_{k = 0}^{n}{1 \over k!}\,{\expo{-k} \over
\bracks{n^{-k}\ \times\ \expo{-k}\ \times\ 1}
n^{k}}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\to} & \hspace{5mm}
\bbox[#ffe,15px,border:1px dotted navy]{\ds{\color{#44f}{\sum_{k = 0}^{n}\,\,{1 \over k!}}}} \\ &
\end{align}
