I cannot give you a closed expression for the coefficients; however you can use the fact that
$$(z;q)_n=\frac{(z;q)_\infty}{(q^n z;q)_\infty}$$ to write $(q^n;q)_\infty= \frac{(q;q)_\infty}{(q;q)_{n-1}}$. Now use Euler's pentagonal number theorem
$$ (q;q)_\infty = \sum_{k \in \mathbb{Z}} (-1)^k q^{\frac 1 2 k (3k-1)} $$ as well as the expansion
$$ \frac{1}{(z;q)_n}=\sum_{k=0}^\infty \binom{n+k-1}{k}_q z^k, $$
where $\binom{n}{m}_q$ denotes the $q$-binomial coefficients, which are polynomials in $q$, to obtain
$$(q^n;q)_\infty = \bigg(\sum_{k \in \mathbb{Z}} (-1)^k q^{\frac 1 2 k (3k-1)} \bigg)
\bigg( \sum_{r=0}^\infty \binom{n+r-2}{r}_q q^r\bigg) $$.
This is a series in $q$ with only positive powers. Hope this helps.