Exponential function formula proof How does one arrive at $e^4$ from
$$\sum_{x=0}^{\infty}\frac{ 4^x}{x!}$$
 A: If we denote by
$$f(x)=\sum_{n=0}^\infty \frac{x^n}{n!},$$
then we have $f(0)=1$ and 
$$f'(x)=\sum_{n=0}^\infty \frac{d}{dx}(\frac{x^n}{n!})=\sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!}=f(x),$$
so $f=\exp$: the unique solution of the differential equation $f'=f, f(0)=1$.
A: OK, I think others are taking the wrong view, here - how do you prove that the function is $e^x$. We already have a definition for powers, let's use it.
Let's start by defining
$$
f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}
$$
Now,
$$
\begin{align}
f(x+y) &= \sum_{n=0}^{\infty} \frac{(x+y)^n}{n!}\\
&=\sum_{n=0}^{\infty} \frac{1}{n!}\sum_{i=0}^n \binom{n}{i}x^iy^{n-i}\\
&=\sum_{i=0}^{\infty} x^i \sum_{n=i}^{\infty}\frac{y^{n-i}}{n!}\binom{n}{i}\\
&=\sum_{i=0}^{\infty} x^i \sum_{n=i}^{\infty}\frac{y^{n-i}}{i!(n-i)!}\\
&=\sum_{i=0}^{\infty} \frac{x^i}{i!}\sum_{n=i}^{\infty}\frac{y^{n-i}}{(n-i)!}\\
&=\sum_{i=0}^{\infty} \frac{x^i}{i!}\sum_{n=0}^{\infty}\frac{y^n}{n!}\\
&=f(x)f(y)
\end{align}
$$
Therefore, we are looking at a function of the form $f(x)=a^x$ (because it's a power function - addition becomes multiplication, $f(x+y)=f(x)f(y)$), for some $a$. So what is $a$? For that, we look at f(1).
$$
f(1) = \sum_{n=0}^\infty \frac{1}{n!} = a
$$
Now, $a$ isn't a rational number, and if we pretend we don't know $e$, then it's a number we've never seen before.
And so, we have $f(x)=e^x$, because the constant, $a$, happens to be Euler's number, $e$.
And if you want to look at the specific case of $e^4$, then notice that $f(1)=e$, and $f(2n)=f(n+n)=f(n)f(n)$, so $f(2)=f(1)^2=e^2$, and $f(4)=f(2)^2=e^4$.
