$Ric$ flat metric on $S^n$.

Question

Suppose $\mathbb{R}\times S^{n}$ admits a complete riemannian metric $g$ such that $Ric_g = 0$. Prove that this metric $g$ induces a metric $\tilde g$ on $S^{n}$ such that $Rig_{\tilde g} = 0$. So far, I was able to prove the following. The manifold $M := \mathbb{R}\times S^{n}$ has two ends. Therefore, any complete metric has a line. By the splitting theorem of Cheeger-Gromoll, $M$ is a riemannian product $M = \mathbb{R} \times N$ for some manifold $N$. One can prove that $N$ has to be a compact orientable manifold and homotopic to the sphere. The splitting of the metric induces a metric $\tilde g$ on $N$ satisfying $Rig_{\tilde g} = 0$. If I'm not mistaken, $N$ is diffeomorphic to $S^n$ is a consequence of the Poincaré conjecture, but there must be a elementary way to prove this. Any hint is appreciated.