According to the pentagonal number theorem:
$$\prod_{n=1}^{\infty} (1-q^{n}) = \sum_{k=-\infty}^{\infty} (-1)^{k}q^{k(3k-1)/2}$$
Now the reciprocal of this has the partition numbers $p(k)$ in its power series representation:
$$\prod_{n=1}^{\infty} \frac{1}{1-q^{n}} = \sum_{k=0}^{\infty} p(k)q^{k}$$
What strikes me about this pair is that there is information about the partitions encoded in the generalized pentagonal numbers.
Are there other pairs of $q$ products and reciprocals for which there is a lot of curiosity about the reciprocal because some number-theoretic information is encoded in the coefficients of the power series representation, and the non-reciprocated power series representation has fairly straightforward exponents?
If the answer to 1 is yes, then also: how do those exponents relate to the information being encoded by the coefficients of the power series representation of the reciprocal?
The number of partitions of $n$ into distinct parts is equal to the number of partitions of $n$ into odd parts:
$$\sum_{n=0}^\infty p_d(n)q^n = (-q;q)_\infty = \frac{(q^2;q^2)_\infty}{(q;q)_\infty} = \frac{1}{(q;q^2)_\infty} = \sum_{n=0}p_o(n)q^n.$$
You can produce many like this, and it is good practice to prove them.
Most incredible and beautiful is Ramanujan's Identity: $$\sum_{n=0}^\infty p(5n+4) q^n = 5\frac{(q^5;q)^5}{(q;q)^6}.$$ This proves that $p(5n+4) \equiv 0 \pmod 5$!