Rewrite limit $\lim_{h \to 0}\frac{e^h - 1}{h}=1$ to $\lim_{h \to 0}\left(1 + h\right)^{\frac{1}{h}}=e$

Question

Rewrite limit $\lim_{h \to 0}\frac{e^h - 1}{h}=1$ to $\lim_{h \to 0}\left(1 + h\right)^{\frac{1}{h}}=e$

Is there any (preferably simple) method of rewriting $\lim_{h \to 0}\frac{e^h - 1}{h}=1$ into $\lim_{h \to 0}\left(1 + h\right)^{\frac{1}{h}}=e$?

You cannot use regular algebraic manipulations as it would create false logical implications:

$$\lim_{h\to 0}1+h=\lim_{h\to 0}e^h \kern.6em\not\kern -.6em \implies \lim_{h\to 0}(1+h)^{1/h}=e$$

Disclaimer:

I am NOT trying to prove the derivative of the exponential function here, my question is a different problem that I would like to address using ONLY limit laws, if possible. This question was already addressed a few weeks earlier, but the explanation here seemed somewhat very difficult and out-of-reach of many calculus students like me.

Answer 1

By a change of variable $h=\log (1+x) \to 0$ as $x\to 0$ we have

$$\lim_{h \to 0}\frac{e^h - 1}{h}=\lim_{x \to 0}\frac{e^{\log (1+x)} - 1}{\log (1+x)}=\lim_{x \to 0}\frac{x}{\log (1+x)}\color{blue}{\cdot\frac{\frac1x}{\frac1x}}=\lim_{x\to 0}\frac1{\color{blue}{\frac1x}\log (1+x)}=\lim_{x \to 0}\frac{1}{\log (1+x)^\color{blue}{\frac1x}}=1$$

which implies

$$\lim_{x \to 0}(1+x)^\frac1x=e$$