Largest power of a prime dividing $q^m-1$

Question

For positive integers $x$ and $d$ let $v_d(x)$ be the largest power of $d$ dividing $x$.

Let $q>1$, $m$ a natural number, and $l$ a prime dividing $q-1$. Then I want to show that $$v_l(q^m-1)=v_l(q-1)+v_l(m)$$ with the exception of $l=2$, $v_2(q-1)=1$ and $v_2(m) \geq 1$

The source I found this in refers to only to "van der Waerden" with no further details.

Answer 1

This is also known as Lifting the Exponent Lemma (LTE).

Hint: When $l$ is odd, use binomial theorem to prove that $v_l(q^l-1)=v_l(q-1)+1$ and $v_l(q^m-1)=v_l(q-1)$ for $l \nmid m$. Now induct on $v_l(m)$.

Where does this fail for $l=2$? Note that $v_2(q^2-1)=v_2(q-1)+1$ only holds for $v_2(q-1) \geq 2$.