Show that if $ f'(0)=1, f'(x)=e^x $ with the law of exponents I'm really stuck on this one. Show that, if $f'(0)=1$ is known, one can deduce
$
f'(x)=e^x
$ with the aid of the law of exponents.
I started here and assumed that I should go towards L'hospital's rule but I get stuck after.

$\lim \limits_{x \to 0} \frac{f(x) - f(0)}{x}
$

The wording exactly: 
Let $f(x)=e^x$. Explain why $ f'(0)= \lim \limits_{x \to 0} \frac{e^x - 1}{x}$. Show that, if $f'(0)=1$ is known, one can deduce $f'(x)=e^x$ with the aid of the laws of exponents.

The first part was trivially easy but the way the problem is worded makes it seem like two separate sets of facts to deduce from.
EDIT: To clarify for you, the problem is asking you to derive the fact that the derivative of $e^x$ is $e^x$ from the basic laws of exponents, rather than as an inherent rule of derivatives.

Given $f(x) = e^x$ we see that $f'(a) = \lim\limits_{x \to a} \dfrac{e^x - e^a}{x-a}$. Therefore $f'(0) = \lim\limits_{x \to 0} \dfrac{e^x - e^0}{x-0} = \lim\limits_{x \to 0}\dfrac{e^x - 1}{x}$. Given that $f'(0) = 1$, we naturally see $\lim\limits_{x \to 0}\dfrac{e^x - 1}{x} = 1$.

So we now know $f(x) = e^x$ and $f'(0) = \lim\limits_{x \to 0}\dfrac{e^x - 1}{x} = 1$. Use these facts to show $f'(x) = e^x$. This can be done as follows:

The general form of a derivative $f'(x) = \lim\limits_{h \to 0}\dfrac{f(x+h) - f(x)}{h}$. In this case, $f'(x) = \lim\limits_{h \to 0}\dfrac{e^{x+h}-e^x}{h}$. By the properties of exponents, $f'(x) = \lim\limits_{h \to 0}\dfrac{e^xe^h-e^x}{h}=\lim\limits_{h \to 0}\dfrac{e^x(e^h - 1)}{h}$.

Because the limit is with respect to $h$ (not $x$), $e^x$ is a constant, and can be pulled out of the limit such that $f'(x) = e^x \lim\limits_{h \to 0} \dfrac{e^h - 1}{h}$. Substituting $h$ for $x$ in the given fact that $\lim\limits_{x \to 0}\dfrac{e^x - 1}{x} = 1$ we see $f'(x) = e^x (1) = e^x$.
There exist an infinite number of functions whose derivative, at x= 0, is 1. It is certainly NOT true that the derivative must be f'(x)= e^x.