Show that $\det:\mathbb{R}^{n\times m}\rightarrow\mathbb{R}, M \mapsto \det(M)$ is differentiable (without having to calculate its derivative.
Show that its inverse is differentiable as well, i.e. for $\text{GL}(n)=\{M\in\mathbb{R}^{n\times m}:\det(M)\neq 0\}$ we have $\text{GL}(n)\rightarrow \text{GL}(n), M\mapsto M^{-1}$ is differentiable.
Am I correct here that the correct procedure is to express the function as a polynomial making use of the Leibniz formula, and then argue that this function is infinitely differentiable as a composition of differentiable functions? Or am I missing a step here, or is possibly expressing it as a polynomial not allowed (or rather I would have to prove its polynomial representation first)?
This one I'm lost on and wouldn't know how to do. It probably involves Cramer's rule in some fashion and the Jacobian matrix and determinant, but I would be incapable of writing this up I think. Does anybody know where I could look at a proof of this?
All of your ideas are correct.
By definition (one of them, at least), we have that $\det((a_{ij})_{i,j=1}^n)= \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{i=1}^na_{i\sigma(i)}$ is a polynomial in the $n^2$ variables $a_{ij}$. So, we have that $\det$ is of class $\mathscr{C}^\infty$.
By the formula $A^{-1} = \det(A)^{-1} \operatorname{Adj}(A)$, we have that the entries of $A^{-1}$ are rational functions of the $n^2$ variables $a_{ij}$, so $A\mapsto A^{-1}$ is also of class $\mathscr{C}^\infty$.