If the universal cover of $X$ is contractible, then $\pi_n(X) = {0}$

Question

I know that a covering $p:X\to Y$ is universal if the space $X$ is simply connected.

But I'm not sure how do I prove that If the universal cover of X is contractible, then $\pi_n(X) = {0}$ for $n>1$

I tried searching a little bit about the proof of this claim but everywhere it is just claimed without a proof...

Can someone please explain me how can I prove this?