Proof that $\sum_{n=0}^{\infty}{\frac{x^{n}}{n!}}={\left(\sum_{n=0}^{\infty}{\frac{1}{n!}}\right)}^{x}$?

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, $$ \left(\sum_{n=0}^\infty\frac{1}{n!}\right)^x=\left(e^1\right)^x=e^x=\sum_{n=0}^\infty\frac{x^n}{n!} $$

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.