The number ${101}^{100}-1$ is divisible by

We have,

$\begin{array}{l}{101}^{100}-1\\ =(1+100{)}^{100}-1\\ =\left\{{ }^{100}{C}_0{+}^{100}{C}_1\left(100\right){+}^{100}{C}_2(100{)}^2+\dots {+}^{100}{C}_{100}(100{)}^{100}\right\}-1\\ {=}^{100}{C}_1\left(100\right){+}^{100}{C}_2(100{)}^2+\dots {+}^{100}{C}_{100}(100{)}^{100}\\ =(100{)}^2\left\{1{+}^{100}{C}_2{+}^{100}{C}_3\left(100\right)+\dots {+}^{100}{C}_{100}(100{)}^{98}\right\}\end{array}$

Clearly, RHS is divisible by $(100{)}^2.$

Hence, ${100}^{100}-1$ is divisible by $10000$ and, therefore, by $100$ and $1000$ also.

ALITER We have,

      $\begin{array}{r}(1+x{)}^n-1=nx+\frac{n(n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{3!}{x}^3+\dots \\ +\frac{n(n-1)(n-2)-(n-(n-1\left)\right)}{n!}{x}^n\end{array}$.

$\therefore (1+x{)}^n-1$ is divisible by $nx.$

Hence, ${101}^{100}-1=(1+100{)}^{100}-1$ is divisible by

     $100\times 100=10000.$