The Derivative of $\exp(x)=e^x$ at $x=0$: $\exp'(0)=1$.
Estimation Argument - "Proof":

We'll follow the four step approach to estimate the derivative at $x =0$ by examining $h>0$ chosen close to $0$.

Recall from ELM.NEL.1 that for large values of $n =2^k, (1+\frac1n)^ n \approx e$.

For $h$ in the difference quotient, we choose $h = \frac1n$, for the same large value of $n$. This guarantees the value of  $h>0$ will be close to $0$.
Then at Step 2 of the four step method (subtract) we consider $e^h -1$   which is approximately $((1+ \frac 1n)^n )^h - 1= (1+ \frac 1n ) ^{ n h} - 1$.
Since $n h = 1$, $(1+ \frac 1n)^{ n h} - 1= ((1+ \frac 1n)) - 1= \frac 1n = h$.

Now we go to Step 3 (divide) and we find that when $h=\frac1n$, $\frac {((1+ \frac 1n)^n )^h - 1}h =1$
Thus at Step 4 (Think), with $n$  getting larger, $h$ gets closer to 0 and the estimate (which is $1$) is closer to the derivative.
So the derivative of $f$ at $0$ must be $1$, i.e.,  $f '(0) = 1$
You can explore this further with the following mapping diagram.

CCD.DEest.0