How can I prove $\ e^n=\sum_{k=0}^\infty\frac{n^k}{k!}$ Given 
$$
e=\sum\limits_{k=0}^\infty\frac{1}{k!}
$$
How can I prove 
$$
e^n=\sum\limits_{k=0}^\infty\frac{n^k}{k!}
$$
Can anyone please demostrate the $n=2$ case? Thanks!
 A: We can work by induction. The base case $n=1$ is trivially true. Suppose it is true for $n$, then
$$e^n =\sum_{i=0}^\infty \frac{n^i}{i!}$$
$$e^n e =\sum_{k=0}^\infty \frac{1}{k!} \sum_{i=0}^\infty \frac{n^i}{i!}$$
$${e^{n + 1}} = \sum\limits_{k = 0}^\infty  {\sum\limits_{i = 0}^\infty  {\frac{1}{{k!i!}}} } {n^i} = \sum\limits_{k = 0}^\infty  {\sum\limits_{i = 0}^\infty  {\frac{{\left( {k + i} \right)!}}{{k!i!}}} } \frac{{{n^i}}}{{\left( {k + i} \right)!}}$$
We procede with $m=k+i$ to get 
$${e^{n + 1}} = \sum\limits_{k = 0}^\infty  {\sum\limits_{i = 0}^\infty  {\frac{1}{{k!i!}}} } {n^i} = \sum\limits_{m = 0}^\infty  {\sum\limits_{i = 0}^m {\frac{{m!}}{{\left( {m - i} \right)!i!}}} } \frac{{{n^i}}}{{m!}}$$
$${e^{n + 1}} =\sum\limits_{m = 0}^\infty  {\left( {\sum\limits_{i = 0}^m {{m\choose i}{n^i}} } \right)} \frac{1}{{m!}}$$
$${e^{n + 1}} = \sum\limits_{m = 0}^\infty  {\frac{{{{\left( {n + 1} \right)}^m}}}{{m!}}} $$
Note that we could have left $\infty$ as the upper limit instead of $m$, since the binomial theorem is a special case of the general binomial theorem.
Also note that the change in the index of summation follows the relation established by $k+n=r$.
Since the hypothesis is true for $n=1$, and $k=n \rightarrow k=n+1$, the formula holds for every $n$ a natural number.
A: $$e^2=\sum_{k=0}^\infty\sum_{m=0}^\infty\frac{1}{k!}\frac{1}{m!}=\sum_{k=0}^\infty\sum_{m=0}^\infty\frac{(m+k)!}{k!m!}\frac{1}{(m+k)!}$$
Now, denote $l:=m+k$ and group the terms by $l$
$$e^2=\sum_{l=0}^\infty\sum_{k=0}^l\frac{l!}{k!(l-k)!}\frac{1}{l!}=\sum_{l=0}^\infty\left[\sum_{k=0}^l\binom{l}{k}\right]\frac{1}{l!}=\sum_{l=0}^\infty\left[2^l\right]\frac{1}{l!}$$
A: Use $\displaystyle\frac{d}{dn}e^n=e^n$ and derive the sum, which is equivalent to the sum itself, so it's equivalent to $e^n$.
