Solution
= Solution
When exactly $n-j$ faces have appeared, there are $j$ unseen faces, so each new roll discovers one with probability $j/n$. Let $G_j$ be the number of further rolls needed at that stage. Then
$$
G_j\sim\operatorname{Geom}\left(\frac jn\right),
\qquad j=n,n-1,\ldots,1.
$$
These waiting times concern disjoint successive blocks of independent rolls, so they are independent. Thus the <coupon collector problem> has the decomposition
$$
\boxed{T_n=\sum_{j=1}^nG_j}.
$$