Let $E_n$ be the event of interest (at the $n$-th extraction we get a new coupon type) and let $c_n=1 \cdots m$ be the type of the $n-$th coupon. Then
$$P(E_n) =\sum_{i=1}^m P(E_n | c_n=i) P(c_n=i) = \sum_{i=1}^m (1-p_i)^{n-1} p_i$$
Let $E_n$ be the event of interest (at the $n$-th extraction we get a new coupon type) and let $c_n=1 \cdots m$ be the type of the $n-$th coupon. Then
$$P(E_n) =\sum_{i=1}^m P(E_n | c_n=i) P(c_n=i) = \sum_{i=1}^m (1-p_i)^{n-1} p_i$$