TITLE: $\lim_{n\to \infty} \sum_{i=0}^n \frac {r^i}{i!} = e^r$
QUESTION [5 upvotes]: Prove that $$\lim_{n\to \infty} \sum_{i=0}^n \frac {r^i}{i!} = e^r$$
Using that $\lim_{n\to \infty} \sum_{i=0}^n \frac{1}{i!} = e$ and $\lim_{n \to \infty} (1 + \frac{1}{n})^n = e$ without differentiation, L'Hôpital and Taylor series.   
I have no idea how to do it, any help is appreciated!

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\pars{\sum_{i = 0}^{\infty}{1 \over i!}}^{r} & =
\sum_{i_{1} = 0}^{\infty}\cdots\sum_{i_{r} = 0}^{\infty}
{1 \over i_{1}!\ldots i_{r}!} =
\sum_{i = 0}^{\infty}\sum_{i_{1} = 0}^{\infty}\cdots\sum_{i_{r} = 0}^{\infty}
{\delta_{\sum_{j = 1}^{r}i_{j},i} \over i_{1}!\ldots i_{r}!}
\\[5mm] & =
\sum_{i = 0}^{\infty}{1 \over i!}
\sum_{i_{1} = 0}^{\infty}\cdots\sum_{i_{r} = 0}^{\infty}
{i! \over i_{1}!\ldots i_{r}!}\,\delta_{\sum_{j = 1}^{r}i_{j},i}
\\[5mm] & =
\sum_{i = 0}^{\infty}{1 \over i!}
\bracks{\sum_{i_{1} = 0}^{\infty}\cdots\sum_{i_{r} = 0}^{\infty}
{i \choose i_{1},\ldots,i_{r}}\,1^{i_{1}}\ldots1^{i_{r}}\delta_{\sum_{j = 1}^{r}i_{j},i}}
\\[5mm] & =
\sum_{i = 0}^{\infty}{1 \over i!}\pars{\sum_{i = 1}^{r}1}^{i} =
\sum_{i = 0}^{\infty}{r^{i} \over i!}
\end{align}