Let: : $\displaystyle s_n = \sum_{k \mathop = 0}^n \dfrac {z^k} {k!}$ : $a_n = \left({1 + \dfrac z n}\right)^n$ Then we can express $a_n$ as follows: {{begin-eqn}} {{eqn | l = a_n       | r = \sum_{k \mathop = 0}^n \binom n k \left({\dfrac z n}\right)^k 1^{n-k}       | c = Binomial Theorem: Integral Index }} {{eqn | r = \sum_{k \mathop = 0}^n \dfrac {z^k} {k!} \dfrac {n \paren {n - 1} \cdots \paren {n - k + 1} }{n^k} }} {{eqn | r = 1 + \sum_{k \mathop = 1}^n \dfrac {z^k} {k!} \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n}       | c = by algebraic manipulations }} {{end-eqn}} The limit of the difference between the $k$th terms of $a_n$ and $s_n$ is: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to +\infty} \cmod {\dfrac {z^k} {k!} - \dfrac {z^k} {k!} \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} }       | r = \cmod {\lim_{n \mathop \to +\infty} \paren {\dfrac {z^k} {k!} - \dfrac {z^k} {k!} \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} } }       | c = Modulus of Limit }} {{eqn | r = \cmod {\dfrac {z^k} {k!} \paren {1 - \prod_{j \mathop = 1}^{k - 1} \paren {1 - \lim_{n \mathop \to +\infty} \dfrac j n} } }       | c = Combination Theorem for Limits of Functions }} {{eqn | r = \cmod {\dfrac {z^k} {k!} \paren {1 - 1} }       | c = Sequence of Powers of Reciprocals is Null Sequence }} {{eqn | r = 0 }} {{end-eqn}} To show that $s_n$ and $a_n$ have the same limit, let $\epsilon \in \R_{>0}$. From Tail of Convergent Series tends to Zero, it follows that we can find $M \in \N$ such that for all $m \ge M$: :$\displaystyle \sum_{k \mathop = m}^n \cmod {\dfrac {z^k} {k!} } < \dfrac \epsilon 2$ For all $k \in \left\{ {0, 1, \ldots, M - 1}\right\}$, we can find $N_k \in \N$ such that for all $n \ge N_k$: :$\displaystyle \cmod{ \dfrac {z^k} {k!} - \dfrac {z^k} {k!} \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} } < \dfrac \epsilon {2 M}$ Then for all $n \ge \max \paren {M, N_0, N_1, \ldots, N_{M - 1} }$, we have: {{begin-eqn}} {{eqn | l = \cmod {s_n - a_n}       | r = \cmod {\sum_{k \mathop = 1}^{M - 1} \dfrac {z^k} {k!} \paren {1 - \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} } + \sum_{k \mathop = M}^n \dfrac {z^k} {k!} \paren {1 - \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} } } }} {{eqn | o = \le       | r = \sum_{k \mathop = 1}^{M - 1} \cmod {\dfrac {z^k} {k!} \paren {1 - \prod_{j \mathop = 1}^{k- 1 } \paren {1 - \dfrac j n} } } + \sum_{k \mathop = M}^n \cmod {\dfrac {z^k} {k!} } \cmod {1 - \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} }       | c = Triangle Inequality for Complex Numbers }} {{eqn | o = \le       | r = \sum_{k \mathop = 1}^{M - 1} \cmod {\dfrac {z^k} {k!} \paren {1 - \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} } } + \sum_{k \mathop = M}^n \cmod {\dfrac {z^k} {k!} } }} {{eqn | o = <       | r = \sum_{k \mathop = 1}^{M - 1} \dfrac \epsilon {2 M} + \dfrac \epsilon 2 }} {{eqn | r = \epsilon }} {{end-eqn}} As an Absolutely Convergent Series is Convergent, $\sequence {s_n}$ converges. Then: {{begin-eqn}} {{eqn | l = 0       | r = \lim_{n \mathop \to +\infty} \cmod {s_n - a_n} }} {{eqn | r = \lim_{n \mathop \to +\infty} s_n - \lim_{n \mathop \to +\infty} a_n       | c = Combination Theorem for Limits of Functions }} {{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {z^k} {k!} - \lim_{n \mathop \to +\infty} \paren {1 + \dfrac z n}^n }} {{end-eqn}} The result follows.