2 of 3 Minor Edit. Fixed Row Alignment.

Hint: For $h\neq0$, we have $~\displaystyle\int_1^xt^{~h-1}~dt=\bigg[\frac{t^h}h\bigg]_1^x=\frac{x^h-1}h.~$ For $h=0$, we have $~\displaystyle\int_1^x\frac{dt}t=$

$=\displaystyle\lim_{h\to0}\frac{x^h-1}h.~$ At the same time, $~\Big(a^x\Big)'=\displaystyle\lim_{h\to0}\dfrac{a^{x+h}-a^x}h=a^x~\lim_{h\to0}\dfrac{a^h-1}h.~$ Do you notice

anything “suspicious” ? ;-) Let e be the number for which this limit is $1$, and let $\ln=\log_{~\large e}.$ Then it

follows that $~\Big(a^x\Big)'=\Big(e^{x\ln a}\Big)'=(x\ln a)'\cdot e^{x\ln a}\cdot1=a^x\ln a.~$ Thus, $~\displaystyle\int_1^x\frac{dt}t=\ln x,$ which

for $x=e,~$ yields $~\displaystyle\int_1^{\large e}\frac{dt}t=1.~$ Now, let us inspect the numbers $u(h)$ for which $~\displaystyle\int_1^{u(h)}t^{h-1}~dt$

$=1.~$ Integrating, we have $\dfrac{u^h-1}h=1\iff u=\sqrt[^h]{1+h}.~$ Letting $h\to0,~$ we have $e=u(0)$

$=\displaystyle\lim_{h\to0}(1+h)^{^\tfrac1h}=\lim_{n\to\infty}\bigg(1+\dfrac1n\bigg)^n$. Hope this helps.