Let $\sigma$ be a $n$-simplex and $K$ the simplicial complex generated by $\sigma$ and its faces. Show that for $0<q<n$, $H_q(K^q;\mathbb{F})$ is a $\mathbb{F}$-vector space of dimension $\binom{n}{q+1}$.
By $K^q$ I mean the $q$-skeleton of $K$, i.e., the simplicial complex formed by the simplices of $K$ of dimension equal or less than $q$.
From another exercise, I've got that $H_q(K^q;\mathbb{F})=H_q(\partial K;\mathbb{F})\oplus \mathrm{Im}{\partial_{q+1}}$ where $\partial K$ denotes the boundary of the simplicial complex (in a topological sense, in this case it is the simplicial complex formed by the $n-1$-simplices of $K$) and $\partial_{q+1}$ is the boundary map of the chain complex of $\partial K$. For $q=n-1$, $\mathrm{Im}{\partial_{q+1}}=0$ because there are no $n$-simplices in $\partial K$. In this case, $H_{n-1}(\partial K;\mathbb{F})\cong\mathbb{F}$, so the dimension is $1=\binom{n}{n-1+1}$.
I'm stuck in the general case. For other values of $q$, $H_q(\partial K;\mathbb{F})=0$, so the dimension of $H_q(K^q;\mathbb{F})$ equals the dimension of $\mathrm{Im}{\partial_{q+1}}$. Since $\sigma$ is an $n$-simplex, I can choose a $q+1$ simplex by choosing $q+1$ vertices from the list of $n+1$ vertices of $\sigma$, so there is a total of $\binom{n+1}{q+1}$ simplices of this dimension. But some of their images are going to be linearly dependent. Here's where I don't know how to continue, I haven't been able to proove that I can choose exactly $\binom{n}{q+1}$ simplices of dimension $q+1$ whose images are linearly independent.
Consider the homology of $K$ itself. It's well-known that $H_j(K,F)=0$ except $H_0(K,F)=F$. Now the complex $K^q$ is the same as $K$ in dimensions less than $q$, so $H_j(K^q,F)=H_j(K,F)$ for $j<q$. Obviously $H_j(K^q,F)=0$ for $j>q$. Consider the Euler characteristic of $K^q$. It is $\chi(K^q)=1+(-1)^q\dim H_q(K^q,F)$. But $\chi(K^q)$ is the alternating sum of the numbers of simplices in $K^q$ so can be calculated. Then $\dim H_q(K^q,F)$ drops out.