I am reading an article that says that, if $l$ is a prime dividing $q-1$, then $l$ divides $\frac{q^l-1}{q-1}$ and $l^2$ does not divide $\frac{q^l-1}{q-1}$ (it is clear that $l$ divides $\frac{q^l-1}{q-1}$). They simply say that
$\frac{q^l -1}{q-1} \equiv \frac{1+(q-1))^l}{q-1} \equiv l + (q-1)\frac{l(l-1)}{2} \mod l^2$,
and the result follows. I just can't see the last equality. I can see that
$\frac{q^l -1}{q-1} = (q-1)((q-1)^{l-1} + {{l}\choose{1}}(q-1)^{l-3} + \ldots + {{l}\choose{l-2}}) $,
we have that ${{l}\choose{l-2}} = \frac{l(l-1)}{2}$, and $q-1 \equiv 0 \mod l$ implies that $l(q-1) \equiv 0 \mod l^2$, therefore
$\frac{q^l -1}{q-1} \equiv (q-1)((q-1)^{l-2} + {{l}\choose{1}}(q-1)^{l-3} + \ldots + {{l}\choose{l-2}}) \equiv (q-1)((q-1)^{l-2} + {{l}\choose{l-1}}) \mod l^2,$
so, to get the result stated above, one must show $(q-1)^{l-2} \equiv 0 \mod l^2$, but I can't see this.
This is an article related with arithmetic of finite fields, thus $q$ is a prime power (not a power of $l$), but I don't think one needs this fact to prove this.
Start with: $$\frac{q^l-1}{q-1}=q^{l-1}+q^{l-2}+\cdots q^1+q^0$$
Now do your substitution $q=(q-1)+1$ and and using $l^2\mid (q-1)^2$ you get:
$$q^{k}\equiv 1 + k(q-1)\pmod{l^2}$$
Now sum $k=0$ to $k=l-1$ you and get:
$$\begin{align}\frac{q^l-1}{q-1}&\equiv \sum_{k=0}^{l-1} \left(1+k(q-1)\right)\pmod{l^2}\\ &=l + (q-1)\frac{l(l-1)}{2} \end{align}$$