Proof of existence of $a$ in $\lim_{h\rightarrow0}\frac{a^h-1}{h}=1$ In deriving the derivative of function $f(x)=\exp(x)$, it is often pointed out that in the general case of $f_a(x)=a^x$ the following expression can be deduced from the definition of the derivative:
$$
\frac{d}{dx}a^x = a^x\cdot\left(\lim_{h\rightarrow0}\frac{a^h-1}{h}\right)
$$
with $e$ defined as the real number for which the above limit term is equal to 1.
However, this seems unsatisfactory to me as the existence of a real number $a$ satisfying the equation $\lim_{h\rightarrow0}\frac{a^h-1}{h}=1$ is not established.
Can somebody please point me to a proof of the existence of such a number or provide it if possible?
Thanks.

One possible approach is to define $e$ as the limit of the sequence 
$$(1+\frac{1}{n})^n$$
It takes a bit of work to show that the above sequence is increasing, and not hard to show that is bounded above by $3$. 
With some more work one shows that 
$$(1+\frac{1}{n})^{n+1}$$
is decreasing. They both have the same limit, which one defines to be the number $e$.
From here one checks that
$$e^{\frac{1}{n+1}}<1+\frac{1}{n}<e^{\frac{1}{n}}$$
and this implies
$$\lim_{x\to 0_{+}}\frac{e^x-1}{x}=1$$
and from this 
$$\lim_{x\to 0}\frac{e^x-1}{x}=1$$
