Arithmetic of integers: divisibility, modular congruence.

Show that $70$ divide $101^{6n} - 1$ for all $n$ natural numbers.

I tried to show that $101^{6n}$$ \equiv 1$ mod $70$.

Thanks for all. I got it.

Note that $70$ $=$ $7$.$5$.$2$

As $101^{ϕ(7)}$ $\equiv 1$ mod $7$, so $101^{6n}$ $\equiv 1$ mod $7$ (Euler Theorem)

and $101^{ϕ(2)}$ $\equiv 1$ mod $2$, so $101^{6n}$ $\equiv 1$ mod $2$ (Euler Theorem)

and $101$ $\equiv 1$ mod $5$, so $101^{6n}$ $\equiv 1$ mod $5$

Therefore, $101^{6n}$ $\equiv 1$ mod $70$.