How Very Odd

we may use $a^2-b^2=(a-b)(a+b)$ iteratively to get

$m^{2^{1000}}-1=(m-1)\prod_{j=0}^{999}(m^{2^j}+1)$

then two cases are considered, knowing that either $m-1$ or $m+1$ is divisible by $4$ . If $4|m-1 \implies m \equiv 1 \ (mod \ 4)$ then $4 \nmid m^{2^j}+1$ for all $j$ . So, we have a total of $1001$ terms that are all divisible by $2$ , except for $m-1$ that is divisible by $4$ and the whole expression would be divisible by $2^{1002}$ .

Same reasoning can be applied to the case $4|m+1$ to deduce the whole expression would be divisible by $2^{1002}$ .

The LTE lemma says that $\nu_2\left(m^{2^{1000}}-1 \right) = \nu_2(m-1) + \nu_2(2^{1000}) + \nu_2(m+1) -1,$ so $\nu_2\left( m^{2^{1000}} - 1 \right) = 999 + \nu_2(m^2-1).$

Now $m^2-1$ is always divisible by $8,$ so $\nu_2(m^2-1) \ge 3,$ so $\nu_2\left( m^{2^{1000}} - 1\right) \ge 1002,$ and if $m=3$ then equality holds.