Is it true that $\pi_n(X^{n-1})=0$ for $n>1$ if $X$ is $K(G,1)$ space?

Question

In this post, Joe Johnson 126 mentioned the above fact, which I'm skeptical of. It is well-known that $\pi_n(X^{n+1})=\pi_n(X)$, but being a $K(G,1)$ space doesn't seem to imply the identity in the title. Also, he alluded that using Hurewicz one obtains $\pi_n(X^n,X^{n-1})\simeq\tilde{H}_n(X^n/X^{n-1})$. But this doesn't necessarily follow according to Hatcher's relative version of Hurewicz theorem, as $X^{n-1}$ isn't simply connected.

Answer 1

This isn't true at all. For instance, let $X=\Delta^3$. Then $X^2=\partial\Delta^3\cong S^2$, so $\pi_3(X^2)=\mathbb{Z}$. But $X$ is contractible, so it is a $K(G,1)$ (with $G$ trivial).