Proof that $\sum_{n=0}^{\infty}{\frac{x^{n}}{n!}}={\left(\sum_{n=0}^{\infty}{\frac{1}{n!}}\right)}^{x}$? The thing is that:
$$
 e^x=\displaystyle\sum_{ n=0 }^{ \infty }\frac{ x^n }{ n! }
$$
$$
e_1=\sqrt[ x ]{ \displaystyle\sum_{ n=0 }^{ \infty }\frac{ x^n }{ n! } }
$$
and
$$
e^1=\displaystyle\sum_{ n=0 }^{ \infty }\frac{ 1^n }{ n! }
$$
$$
e_2=\displaystyle\sum_{ n=0 }^{ \infty }\frac{ 1 }{ n! }
$$
If we equate this two expressions, we get
$$
e_1=e_2
$$
which is equal to
$$
\sqrt[ x ]{ \displaystyle\sum_{ n=0 }^{ \infty }\frac{ x^n }{ n! } }=\displaystyle\sum_{ n=0 }^{ \infty }\frac{ 1 }{ n! }
$$
$$
\displaystyle\sum_{ n=0 }^{ \infty }\frac{ x^n }{ n! }=\left(\displaystyle\sum_{ n=0 }^{ \infty }\frac{ 1 }{ n! }\right)^x
$$
But how can I prove that? Is it already proven?
 A: In this answer, it is shown that using the definition
$$
e=\lim_{n\to\infty}\left(1+\frac1n\right)^n
$$
we get
$$
e^x=\sum_{n=0}^\infty\frac{x^n}{n!}
$$
Therefore,
$$
\left(\sum_{n=0}^\infty\frac{1}{n!}\right)^x=\left(e^1\right)^x=e^x=\sum_{n=0}^\infty\frac{x^n}{n!}
$$
A: For any $x,y \in \mathbb{R}$, the family $(\frac{x^ny^m}{n!m!})$ indexed by $(n,m) \in \mathbb{N}^2$ is summable, so by reordering the terms we get that :
$$ \left(\sum_{n \in \mathbb{N}} \frac{x^n}{n!}\right)\left(\sum_{m \in \mathbb{N}} \frac{y^m}{m!}\right)
= \sum_{(n,m) \in \mathbb{N}^2} \frac{x^n y^m}{n! m!}
= \sum_{p \in  \mathbb{N}} \left( \sum_{n+m=p} \frac{x^n y^m}{n! m!} \right) \\
= \sum_{p \in  \mathbb{N}} \frac1{p!}\left( \sum_{n+m=p} \binom p n x^n y^m \right) 
= \sum_{p \in  \mathbb{N}} \frac{(x+y)^p}{p!}
$$
Therefore the function $\exp : x \mapsto \sum \frac{x^n}{n!}$ is a group morphism from $(\mathbb{R},+)$ to $(\mathbb{R}^*,*)$.
But it also is continuous, so it is a continuous morphism from $(\mathbb{R},+)$ to $(\mathbb{R}_+^*,*)$ 
Now, exponentiation $x^y$, is usually defined on $(\mathbb{R}_+^* \times \mathbb{R})$ by the property that forall $x \in \mathbb{R}_+^*$, the function $y \mapsto x^y$ is the unique continuous morphism from $(\mathbb{R},+)$ to $(\mathbb{R}_+^*,*)$ such that $x^1 = x$.
Indeed, for any $x \in \mathbb{R}_+^*$, there is only one way to define $x^n$ for $n \in \mathbb{Z}$ such that it is a group morphism. 
Since for any positive integer $q$, the map $x \mapsto x^q$ is a bijection from $\mathbb{R}_+^*$ to itself, we have to define $x^{p/q}$ for $(p/q) \in \mathbb{Q}$ as the unique number in $\mathbb{R}_+^*$ such that $(x^{p/q})^q = x^p$. 
And finally, assuming the function on $\mathbb{Q}$ we have defined so far is continuous, since $\mathbb{Q}$ is dense in $\mathbb{R}$, there is only one way to extend this to a continuous morphism defined on all of $\mathbb{R}$
Since $\exp$ is a continuous group morphism sending $1$ to $\exp(1)$, we can conclude that it is the continuous group morphism sending $1$ to $\exp(1)$, that is, we can take $\exp(1)^y = \exp(y)$ as a definition of $\exp(1)^y$.
More generally, for any $y \in \mathbb{R}$, $x \mapsto \exp(x)^y$ and $x \mapsto \exp(xy)$ are both continuous group morphisms sending $1$ to $\exp(y)$, so they have to coincide.
A: Assume that we work with the following definitions: 
$Exp(x):=\sum_{n=0}^\infty\frac{x^n}{n!} $ and $e:=\sum_{n=0}^\infty\frac{1}{n!}$. Then what you want to show is that $Exp(x)=e^x$ all real $x$. 
To start with it is convenient to show that $Exp(x+y)=Exp(x)Exp(y)$ for any $x,y\in \mathbb{R}$. As pointed out in the comments this can be done using the fact that the cauchy product of two series converges if one of them is absolutely convergent. 
By induction we can extend the formula so that $Exp(x_1+…+x_n)=Exp(x_1)\dots Exp(x_n)$ for $n\in \mathbb{N}$ and $x_1,…,x_n \in\mathbb{R}$. (Try to show this). 
Secondly we observe that $Exp(1)=e$, and hence by above it follows that $Exp(n)=e^n$ all $n\in\mathbb{N}$. 
Now the leap to showing this for all rational $q=m/n$ $m,n$ positive  is not that big, just observe that $Exp(q)^n=Exp(qn)=Exp(m)=e^m$ all $m,n\in\mathbb{N}$ and hence $Exp(q)=(e^m)^{1/n}$. If we define $a^q=(a^m)^{1/n}$ all $q\in\mathbb{Q}$, then we have shown that $Exp(q)=e^q$ all $q\in\mathbb{Q^+}$. Thirdly we can extend this result to all of $\mathbb{Q}$ by noting that $Exp(x)Exp(-x)=Exp(x-x)=Exp(0)=1$ all $x\in\mathbb{R}$. 
It remains to show your equality for all real $x$. If we define exponentiation by an arbitrary real number $y$ by the following: $$a^y:=sup\space a^q$$ where the supremum is taken over all rational $q$ less than $y$, then we can quite smoothly show your equality. To do that we first have to establish that $Exp(x)$ is continuous and monotone on the real axis. (I leave it for you to do that ).
Lastly we have $$Exp(x)=sup_{q<x}\space Exp(q)=sup_{q<x}\space e^q=e^x$$ 
where the first equality follows by continuity and monotonicity of $Exp(x)$ and the second equality follows by the results established above. The last equality follows by how we defined exponentiation for $x\in\mathbb{R}$
Please let me know if you want me to clarify anything. I shall also add that what I have written above resembles very much the exposition in Baby Rudin, so it is far from original. In case you do not have the book, it may be useful. 
