Consider the projective space $P(\mathbb{F}^{n+1}_q)$, the projective space constructed over $\mathbb{F}^{n+1}_q$, where $q$ is prime and $n \in \mathbb{N}$.
How many points does it have? And how many straight lines? I've already figured it out for $q=2$, but I don't know how to generalise it to every prime number $q$.
In general, the number of $m$-dimensional projective subspaces in $PG(n,q)$ is $\frac{(q^{n+1}-1)(q^{n+1}-q)\dots(q^{n+1}-q^m)}{(q^{m+1}-1)(q^{m+1}-q)\dots(q^{m+1}-q^m)}.$
If $m=0$, we get that the number of points is in fact $\frac{q^{n+1}-1}{q-1}$,
if $m=1$, we get that the number of lines is $\frac{(q^n+q^{n-1}+\dots+q+1)(q^{n-1}+\dots+q+1)}{q+1}$.