$\lim_{n\to \infty} \sum_{i=0}^n \frac {r^i}{i!} = e^r$ 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!
 A: Firstly you should show that 
Lemma 1:
$\left(\left(1+\dfrac k n\right)^n\right)_n$,this sequence is monotonous increasing and has a upperbound $e^k$.
Proof:
Let $k\ge x$ be a natural number;
$$1+\dfrac xn\le 1+\dfrac kn$$$$\Rightarrow$$$$\left(1+\dfrac k n\right)^n\le \left(1+\dfrac k {kn}\right)^{kn}=\left(1+\dfrac k n\right)^{kn}\le e^k\\ \Box.$$
Lemma 2:
Let $q$ be $\in\mathbb N$,so limit will be $\lim\limits_{n\to\infty}(1+q/n)^n=e^q$
Proof:
From "lemma 1" we can write;
$$\lim\limits_{n\to\infty}\left(1+\dfrac q n\right)^n\le e^q$$ Therefore we should show the following equation;
$$\lim\limits_{n\to\infty}\left(1+\dfrac q n\right)^{n/q}\le e$$
Start with $\left(1+\dfrac q n\right)^n$ and consequence will be like following;
$$\lim\limits_{n\to\infty}\left(1+\dfrac q n\right)^n=\lim\limits_{n\to\infty}\left[\left(1+\dfrac q n\right)^{n/q}\right]^q=\left[\lim\limits_{n\to\infty}\left(1+\dfrac q n\right)^{n/q}\right]^q=\left[\lim\limits_{u\to\infty}\left(1+\dfrac 1 u\right)^{u}\right]^q=e^q\\ \Box.$$
If we took $q\in\mathbb Q$, proof would be like that simply.After all of these we can ultimately say;
$$\boxed{\forall q \in\mathbb Q \\ \lim\limits_{n\to\infty}\left(1+\dfrac q n\right)^n=e^q}$$
On the other hands, we need to prove that $e^q$  equals to $\lim\limits_{n\to\infty}\left(1+\dfrac q n\right)^n$  and $\displaystyle\sum_{i=0}^\infty q^i/i!$ at the same time.
Lemma 3: 
If $0\le a_i\le x$, then;
$$(x-a_1)(x-a_2)...(x-a_k)\ge x^k-x^{k-1}(a_1+a_2+...+a_k)$$
Proof:(Induction)
If
$$(x-a_1)(x-a_2)...(x-a_k)\ge x^k-x^{k-1}(a_1+a_2+...+a_k)$$ is true.
$$(x-a_1)(x-a_2)...(x-a_k)(x-a_{k+1})\ge x^{k+1}-x^{k}(a_1+a_2+...+a_k)$$ should be true.
$$(x-a_1)(x-a_2)...(x-a_k)(x-a_{k+1})\ge \left(x^k-x^{k-1}(a_1+a_2+...+a_k)\right)(x-a_{k+1})\\=x^{k+1}-(a_1+a_2+...+a_k+a_{k+1})x^k+(a_1+a_2+...+a_k)x^{k-1}a_{k+1}\\ \ge x^{k+1}-x^k(a_1+a_2+...+a_{k+1})\\\Box.$$
Lemma 4:
For $\forall i (2\le i\le n)$;$$0\le \dfrac1{i!}-\dbinom{n}{i}\dfrac1{n^i}\le \dfrac1{2n}\dfrac{1}{(i-2)!}$$
Proof
$$\dfrac1{i!}-\dbinom{n}{i}\dfrac1{n^i}=\dfrac1{i!}-\dfrac{n!}{(n-i)!i!n^i}=\dfrac1{i!}\left(1-\dfrac{n!}{(n-i)!n^i}\right)\\=\dfrac1{i!}\dfrac{n^{i-1}-(n-1)(n-2)...(n-i+1)}{n^{i-1}}$$Use here "Lemma 3"$$\dfrac1{i!}\dfrac{n^{i-1}-(n-1)(n-2)...(n-i+1)}{n^{i-1}}\\\le \dfrac{1}{i!}\dfrac{n^{i-1}-(n^{i-1}-(1+2+3+4+...+(i-1))n^{i-2})}{n^{i-1}}\\=\dfrac{(i-1)i}{2i!n}=\dfrac{1}{2n}\dfrac{1}{(i-2)!}$$
Theorem:
$$\displaystyle\sum_{i=0}^\infty \dfrac{x^i}{i!}=\lim\limits_{n\to\infty}\left(1+\dfrac xn\right)^n=e^x$$
Proof:
$$\left|\displaystyle\sum_{i=0}^n \dfrac{x^i}{i!}-\left(1+\dfrac xn\right)^n\right|\le \displaystyle\sum_{i=2}^n \left|\dfrac{1}{i!}-\dbinom{n}{i}\dfrac1{n^i}\right||x|^i=\displaystyle\sum_{i=2}^n \left(\dfrac{1}{i!}-\dbinom{n}{i}\dfrac1{n^i}\right)|x|^i\longrightarrow\text{Use Lemma 4}\\\le \displaystyle\sum_{i=2}^n \dfrac1{2n}\dfrac1{(i-2)!}|x|^i=\dfrac{|x|^2}{2n}\sum_{i=2}^n \dfrac{|x|^{i-2}}{(i-2)!}\\=\dfrac{|x|^2}{2n}\sum_{j=0}^{n-2} \dfrac{|x|^{j}}{(j)!}\le\dfrac{|x|^2}{2n} e^{|x|}$$
And $\epsilon>0$ has been given.If we choose $2N\epsilon=|x|^2e^{|x|}$, $\forall n>N$ ;
$$\left|\displaystyle\sum_{i=0}^n \dfrac{x^i}{i!}-\left(1+\dfrac xn\right)^n\right|\le ... \le \dfrac{|x|^2}{2n} e^{|x|} \le \dfrac{|x|^2}{2N} e^{|x|}<\epsilon$$$$Q.E.D.\Box$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\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}
A: This is probably not the desired answer, because it uses continuity of power series. But it is still very elementary and I found it worth to present it here.
Define $f:\mathbb R\to\mathbb R$ by $f(x)=\sum_{k=0}^\infty\frac{x^k}{k!}$. Then $f$ is a well-defined continuous function. Moreover, it satisfies the functional equation $f(x+y)=f(x)f(y)$, which can be proved using the Cauchy product formula: Accordingly, for fixed $x,y\in\mathbb R$ we have
\begin{align*}
f(x)f(y)=\sum_{k=0}^\infty\frac{x^k}{k!}\cdot \sum_{k=0}^\infty\frac{y^k}{k!}=
\sum_{n=0}^\infty\sum_{k=0}^n\frac{x^ky^{n-k}}{k!(n-k)!}=
\sum_{n=0}^\infty\frac{1}{n!}\sum_{k=0}^n\binom{n}{k}x^ky^{n-k}=
\sum_{n=0}^\infty\frac{(x+y)^n}{n!}=f(x+y).
\end{align*}
Next, putting $x=y$, we find $f(2x)=f(x)^2$ and by induction $f(nx)=f(x)^n$ for $n\in\mathbb N,x\in\mathbb R$. In particular, $f\geq 0$, and $f(n)=e^n$, since $f(1)=e$. Replacing $x$ by $x/n$ gives $f(x)=f(x/n)^n$, hence $f(x/n)=f(x)^\frac{1}{n}$. Replacing here $x=m$ for $m\in\mathbb N$ yields $f(m/n)=f(m)^\frac{1}{n}=e^\frac{m}{n}$, which means $f(x)=e^x$ for $x\in\mathbb Q,x>0$. Since $1=f(0)=f(x-x)=f(x)f(-x)=e^xf(-x)$ we get $f(-x)=e^{-x}$ for $x\in\mathbb Q,x>0$. Consequently, $f(x)=e^x$ for all $x\in\mathbb Q$. Since $f$ is continuous, $f(x)=e^x$ for all $x\in\mathbb R$.
Note that only the functional equation $f(x+y)=f(x)f(y)$ forces a function $f$ to satisfy $f\geq 0$ and $f(x)=f(1)^x$ for $x\in\mathbb Q$!
