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.
 A: Assuming that basic theorems about limits and the logarithm function are available as well as $\lim_{x\to 0}\left(x+1\right)^{1/x}=e$, we can first prove that:
$$\lim_{y\to 0}\frac{\ln(y+1)}{y}=\lim_{y\to 0}\ln\left(y+1\right)^{1/y}=\ln\left(\lim_{y\to 0}\left(y+1\right)^{1/y}\right)=\ln(e)=1 $$
Switching the limit and the logarithm is possible since $\ln(t)$ is continous. Then
$$\lim_{y\to 0}\frac{\log_a(y+1)}{y}=\frac{1}{\ln(a)}\lim_{y\to 0}\frac{\ln(y+1)}{y}=\frac{1}{\ln(a)} $$
And $y=a^{h}-1\to 0$ as $h\to 0$.
So by limit of a composition of functions,
$$\lim_{h\to 0}\frac{\log_a\left(a^h-1+1\right)}{a^h-1}=\lim_{h\to 0}\frac{h}{a^h-1}=\frac{1}{\ln(a)}\\
\lim_{h\to 0}\frac{a^h-1}{h}=\ln(a)$$
Hence $\lim_{h\to 0}\dfrac{a^h-1}{h}=1$ if and only if $\ln(a)=1\Leftrightarrow a=e$.
A: 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 
$$\lim_{x\to 0}\frac{e^x-1}{x}=1$$
