# 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!

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$$e^2=\sum_{k=0}^\infty\sum_{m=0}^\infty\frac{1}{k!}\frac{1}{m!}=\sum_{k=0}^\infty\sum_{m=0}^\infty\frac{(m+k)!}{k!m!}\frac{1}{(m+k)!}$$

Now, denote $l:=m+k$ and group the terms by $l$

$$e^2=\sum_{l=0}^\infty\sum_{k=0}^l\frac{l!}{k!(l-k)!}\frac{1}{l!}=\sum_{l=0}^\infty\left[\sum_{k=0}^l\binom{l}{k}\right]\frac{1}{l!}=\sum_{l=0}^\infty\left[2^l\right]\frac{1}{l!}$$

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It can be useful to remark that this is true thanks to the absolute convergence of the series that defines $e$. More generally, this follows from a convergence theorem for Cauchy products of series. –  Siminore May 11 '12 at 14:46
The multinomial generalization to all natural $n$ should be clear from this answer. –  anon May 11 '12 at 14:55

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.

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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 $e^n$.

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It seems $n$ is an integer here. –  Pedro Tamaroff Jul 12 '13 at 6:47