Showing that $\mathbb{Q}(\zeta_p, \sqrt[p]{\ell}) = \mathbb{Q}(\zeta_p + \sqrt[p]{\ell})$ for $p,\ell$ primes.

You were very near to the full solution. If $n\neq 0$, you can rewrite your equality as

$$ \sqrt[p]{\ell}=\frac{\zeta_p-\zeta_p^m}{1-\zeta_p^n} $$

So you deduce $\sqrt[p]{\ell} \in {\mathbb Q}(\zeta_p)$, and it is very easy to finish from there.