I attempted this on my own and got a fairly simple solution. However, after reading proofs here and here, I feel like I have massively over simplified the problem. I understand the other solutions, but does my shorter one work?
Proposition. The expected number of cycles of length $k$ in an Erdős–Rényi random graph $G_{n,p}$ is $$\binom nk(k-1)!p^k/2.$$
Proof. A cycle of length $k$ requires $k$ vertices of which there are $\binom nk$ choices. There are $k!$ different cycles/paths we could potentially follow in a cycle on these $k$ vertices. Since the starting point is irrelivant, we can divide by $k$, as is the direction/orientation, giving $(k-1)!/2$ different cycles. For each cycle, we require the $k$ edges between them to be included in the graph, which happens with probability $p^k$. Conclude that the expected number of such cycles is as stated.
Edit: I am only interested in cycles of length $k$, sorry!
