I would like to know if my proof below is correct.
Problem Show that $GL_n(\mathbb{F})$ is a finite group iff $\mathbb{F}$ has a finite number of elements.
Solution
If $\mathbb{F}$ is a finite field and say $\vert \mathbb{F} \vert = m$, then the set of all $n \times n$ matrices, which is a superset of $GL_n(\mathbb{F})$, where the elements comes from $\mathbb{F}$ has cardinality $m^{n^2}$. Hence, $\vert GL_n(\mathbb{F}) \vert < m^{n^2}$ and thereby $GL_n(\mathbb{F})$ is a finite group.
If $GL_n(\mathbb{F})$ is a finite group, note that for every non-zero element $a \in \mathbb{F} -\{0\}$, the matrix $aI_{n \times n}\in GL_n(\mathbb{F})$, where $I_{n \times n}$ is the identity matrix with the multiplicative identity along the diagonal. This means $\vert \mathbb{F} \vert \leq GL_n(\mathbb{F}) + 1$ and thereby $\mathbb{F}$ is a finite field.
Thanks