Suppose $G$ is a finite group with Sylow $p$-subgroups of order $p^n$. I am trying to show that if there are $\geq p+1$ Sylow $p$-subgroups, then the union of all these subgroups has order $\geq p^{n+1}$.
In the case where there are exactly $p+1$ Sylow $p$-subgroups, we can find a homomorphism $G \rightarrow S_{p+1}$, which has kernel $K$ of size $p^{n-1}$ since $(p+1)!$ is divisible by $p$ but not by $p^2$. Then if $P$ is a Sylow $p$-subgroup, the subgroup $PK$ has order power of $p$, hence $|PK| \leq p^n$ and by the order formula for $PK$ we find that $|P \cap K| \geq p^{n-1}$. Therefore $K \leq P$. Since $K$ is contained in every Sylow $p$-subgroup, the intersection of all Sylow $p$-subgroups has order $p^{n-1}$. Thus there are $(p+1)(p^n - p^{n-1}) + p^{n-1} = p^{n+1}$ elements in the union of all Sylow $p$-subgroups.
Now I'm stuck. How can I handle the case where there are $> p+1$ Sylow $p$-subgroups?