A finite groups of $\mathrm{GL}_n(\mathbb C)$ of exponent $m$ necessarily have order $C$ verifying $C\leqslant m^n$ and $n! m^n$ divides $C$, but this condition is not sufficient, for instance $\mathrm{GL}_3(\mathbb C)$ has no subgroup of exponent $7$ and order $42$.

So can we determine fully the possible orders of the subgroups of $\mathrm{GL}_n(\mathbb C)$ of exponent $m$?

~~~~~~~~~~~~OLD QUESTION - Which was obvious, sorry and thanks Pete ~~~~~~~~~~~~~~~~

After some computation and search it seems that finite subgroups of $\mathrm{GL}_3(\mathbb C)$ of exponent $5$ have order $5$, $5^2$ or $5^3$. Similarly, finite subgroups of $\mathrm{GL}_3(\mathbb C)$ of exponent $7$ have order $7$, $7^2$ or $7^3$.

So, for $n >2$ and a prime $p >2$, is it true that finite subgroups of $\mathrm{GL}_n(\mathbb C)$ of exponent $p$ necessarily have order of the form $p^k$? I apologize in advance if this is obvious and I missed it.