Euler or Fermat?

By Fermat's little theorem

$a^{p} \equiv a \pmod p$

Let $a = 100$ and $p = 101$

$100^{101} \equiv 100 \pmod {101}$ $100^{101} - 100 \equiv 0 \pmod {101}$

Thus we get that $\boxed{101 \mid 100^{101} - 100}$ and the answer is $\boxed{\text{Yes}}$

$100^{101}-100 \equiv (-1)^{101}-(-1) \equiv -1+1 \equiv 0 \pmod {101}$ .

Hence $101$ divides $100^{101}-100$ .

$\begin{aligned} &\textcolor{#FFFFFF}{=}100^{101}-100\\ &=100\left(100^{100}-1\right)\\ &=100\left(1\underbrace{0\dots0}_{\times100}-1\right)\\ &=100\left(\underbrace{9\dots9}_{\times100}\right)\\ &=900\left(\underbrace{1\dots1}_{\times100}\right) \end{aligned}$

$\begin{aligned} 101\times11&=1111\\ 101\times110011&=11111111\\ 101\times1100110011&=111111111111\\ &\vdots\\ 101\times11\underbrace{0011\dots0011}_{\times(n-1)}&=\underbrace{1\dots1}_{\times4n}\\ &\vdots\\ 101\times11\underbrace{0011\dots0011}_{\times24}&=\underbrace{1\dots1}_{\times100}\\ 900\left(101\times11\underbrace{0011\dots0011}_{\times24}\right)&=900\left(\underbrace{1\dots1}_{\times100}\right)\\ 101\left(900\times11\underbrace{0011\dots0011}_{\times24}\right)&=100^{101}-100 \end{aligned}$

$100^{101}-100=101k\implies\boxed{101\mid 100^{101}-100}$