Let $M$ be a , connected, compact, $\mathbb{Z}$-orientable topological $n$-manifold. The latter $\mathbb{Z}$-orientable means that $H_n(M; \mathbb{Z})=\mathbb{Z} $ holds.
Let futhermore $M$ be simply connected and it is homotopy equivalent to a suspesion $\Sigma X$ of path connected space $X$. (btw: I'm not sure if we need here this information)
Why in this case $M$ has the same homology groups like $S^n$?
Therefore $H_i(M, \mathbb{Z})= \mathbb{Z}$ if $i=0,n$ and otherwise $H_i(M, \mathbb{Z})=0$.
My attemps:
Let abbreviate $H_i(M; \mathbb{Z}) = H_i(M)$.
Connectedness implies $H_0(M)=\mathbb{Z}$. Since $M$ simple connected Hurewicz implies $0 =\pi_1(M) \twoheadrightarrow H_1(M)$.
Futhermore - since $M$ orientable - $H_n(M)$ and $H_{n-1}(M) =\mathbb{Z}^{k}$ free. But from here I stuck in calculating other homology groups.
I guess I can argue by induction on $i$. The cases $i =0,1,n$ are ok and wlog I can assume that $n >2$ otherwise I finish.
Can anybody help me to get the induction step? I tried using Universal Coefficient Thm reducing the calculation to $H_i(X; F)$ where $F= \mathbb{Z}/p$ field in order to use Poincaré. But also here without success.