= Solution
Because $K$ is a subcomplex, $\partial C_k(K)\subseteq C_{k-1}(K)$, so
$$
\partial(c+C_k(K))=\partial c+C_{k-1}(K)
$$
is well defined and squares to zero. The short exact <sequence> of chain complexes
$$
0\to C_\bullet(K)\to C_\bullet(L)\to C_\bullet(L,K)\to0
$$
gives the long exact <sequence>
$$
\cdots\to H_k(K)\to H_k(L)\to H_k(L,K)
\to H_{k-1}(K)\to\cdots.
$$
Taking $L=\partial\Delta^{n+1}$ and using the contractibility of $\Delta^{n+1}$ gives
$$
H_k(\partial\Delta^{n+1})\cong
\begin{cases}\mathbb Z,&k=0,n,\\0,&\text{otherwise}.
\end{cases}
$$
For $0<k<n$, exactness makes $H_k(K)$ a quotient of a <subgroup> of $H_{k+1}(L,K)$, whose rank is at most
$$
\operatorname{rank}C_{k+1}(L,K)
=\binom{n+2}{k+2}-\#\{(k+1)\text{-simplices of }K\}.
$$
The 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 $H_n(K)=0$.
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