How can I prove that there is a function that is its own derivative?

There are two ways you could show it. The harder route would be to prove the existence and uniqueness theorem for ordinary differential equations, thus showing there exists solutions to $y'=y$.

The more direct way would be to just construct the function $e^x$ and show that it's its own derivative. You would start by defining $$\ln(x) = \int_1^x \frac{1}{t}\, dt$$ and prove that it's a strictly increasing function on $(0,\infty)$ with range $(-\infty, \infty)$. It follows that $\ln(x)$ has an inverse, which we should dub $e^x$. As for finding the derivative of this new and mysterious function: $$y=e^x$$ $$\ln(y)=x$$ Taking the $x$ derivative of both sides, $$\frac{y'}{y} = 1$$ $$\implies y'=y$$ And do show that every function which is its own derivative is a constant multiple of $e^x$, suppose that $f'=f$. Then, noting that $e^x$ is nowhere zero, $$\frac{d}{dx} \frac{f(x)}{e^x} = \frac{f'(x)e^x-f(x)e^x}{(e^x)^2} = \frac{f(x)e^x-f(x)e^x}{(e^x)^2} = 0$$ Therefore, $$\frac{f(x)}{e^x}$$ is constant since it has a connected domain, and so $f(x) = ce^x$ for some $c$.

$f(x) = 0$ is trivially its own derivative, and is of the form $a(b^x)$ for $a=0$ and any positive $b$. That's all we need to solve the problem posed.

An intuitive answer:

For smooth functions and "small" $h$, we have

$$f'(x)\approx\frac{f(x+h)-f(x)}h.$$

Then $f'(x)=f(x)$ yields

$$f(x+h)\approx(1+h)f(x),$$ and $$f(x+2h)\approx(1+h)f(x+h)\approx(1+h)^2f(x),$$ $$\cdots$$ $$f(x+nh)\approx(1+h)^nf(x).$$

Now with $nh=1$,

$$f(x+1)\approx \left(1+\frac1n\right)^n f(x),$$ which should ring a bell.