 Solution

Because $K$ is a subcomplex, $\partial C_k(K)\subseteq C_{k-1}(K)$, sois well defined and squares to zero. The short exact sequence of chain complexesgives the long exact sequenceTaking $L=\partial {\Delta }^{n+1}$ and using the contractibility of ${\Delta }^{n+1}$ givesFor $0<k<n$, exactness makes $H_k(K)$ a quotient of a subgroup of $H_{k+1}(L,K)$, whose rank is at mostThe same stated inequality for $k=n$ follows from $H_n(K)\hookrightarrow H_n(L)\cong \mathbb{Z}$. Finally, an $n$-cycle in the boundary sphere has equal signed coefficients on all $n$-faces. A proper subcomplex omits an $n$-face, so that coefficient and hence every coefficient is zero. Therefore $H_n(K)=0$.