Proof that classifying spaces for discrete groups are the Eilenberg-MacLane spaces

A classifying space $BG$ of a topological group $G$ is the quotient of a weakly contractible space $EG$ by a free action of $G$. The claim is that if $G$ is a discrete group then $EG/G$ is an Eilenberg-MacLane space, so that the fundamental group of $EG/G$ is $G$ and all higher homotopy groups are trivial.

Is it easy to prove such a claim, or does anyone have a reference for this?

Solutions Collecting From Web of "Proof that classifying spaces for discrete groups are the Eilenberg-MacLane spaces"

For any topological group $G$, $EG \to BG$ is a principal $G$-bundle, so by the long exact sequence in homotopy, we have the long exact sequence

$$\dots \to \pi_{n+1}(BG) \to \pi_n(G) \to \pi_n(EG) \to \pi_n(BG) \to \pi_{n-1}(G) \to \dots$$

As $EG$ is weakly contractible, $\pi_n(EG) = 0$ for $n > 0$, so we see that $\pi_{n+1}(BG) \cong \pi_n(G)$. In particular, if $G$ is discrete, $\pi_n(G) = 0$ for $n > 0$ and $\pi_0(G) = G$, so $\pi_n(BG) = 0$ for $n \neq 1$, and $\pi_1(BG) = G$. Therefore, $BG$ is a $K(G, 1)$.

More generally, as $\pi_{n+1}(BG) \cong \pi_n(G)$, if $G$ is itself an Eilenberg-Maclane space, say a $K(H, m)$, then $BG$ is an Eilenberg-Maclane space, namely a $K(H, m + 1)$. We can see this as a generalisation of the above if we consider a discrete topological group $G$ as a $K(G, 0)$. Conversely, if $BG$ is an Eilenberg-Maclane space, $G$ is either an Eilenberg-Maclane space or discrete.