explain the proof for an expression for $e^x$ using the binomial theorem

Question

Below is the proof for an expression for the $e^x$ using the binomial theorem. Could someone explain the second step? where we as n tends to infinity $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \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}\,\,\, \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}} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, \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}{\sim}\,\,\, \bbox[#ffe,15px,border:1px dotted navy]{\ds{\sum_{k = 0}^{n}{1 \over k!}}} \end{align}