number theory - Let $a$ and $b$ be integers $ge 1$. prove that $(2^a-1) | (2^{ab}-1)$.

Let $a$ and $b$ be integers $\ge 1$. prove the following:

$(2^a-1) | (2^{ab}-1)$

My attempt:

$2^{ab}-1=(2^a)^b-1$

$= (2^a-1)((2^a)^{b-1}+(2^a)^{b-2}+...+2^a+1)$

Since $(2^a)^{b-1}+(2^a)^{b-2}+...+2^a+1\in \Bbb{Z}$,
then

$(2^{ab}-1)\equiv 0 \mod (2^a-1)$.

Is that true, please ?