This is an exercise of Advanced Linear Algebra by Roman:
Let $\text{dim}(V)<\infty$. If $\tau, \sigma \in \mathcal{L}(V)$ prove that $\sigma\tau = \iota$ implies that $\tau$ and $\sigma$ are invertible and that $\sigma = p(\tau)$ for some polynomial $p(x) \in F[x]$.
The difficulty in this exercise is that it is stated prior to anything related to eigenvalues or the Caley-Hamilton theorem. I want to confirm that the proof below is correct or, by the contrary, if there is some flaw somewhere.
If $\tau$ is not invertible then there is some $x\in V\setminus\{0\}$ such that $\tau x=0$, but this would imply that $\sigma\tau x=0$, what is not possible, so $\tau$ is invertible. If $\tau$ is invertible then $\tau$ is an automorphism, what imply that $\sigma$ can take any value, so $\sigma$ is invertible by the same reasons than $\tau$.
Then $\sigma=\tau^{-1}$ and we want to show that there is a polynomial $p$ such that $\tau^{-1}=p(\tau)$. Now suppose that $\dim V=n$ and let $v_1,\ldots,v_n$ a basis of $V$, then for each $k\in\{1,\ldots,n\}$ choose the minimal list that contains $\tau^{-1}v_k$ where $\tau^{-1}v_k,\tau^n v_k,\tau^{n+1} v_k,\ldots$ is linearly dependent. Then there are $c_j\in F\setminus\{0\}$, $m_j\ge 0$ and some $1\le n_k< n$ such that $$ c_{-1}\tau^{-1}v_k+\sum_{j=1}^{n_k}c_j\tau^{n+m_j}v_k=0 $$ Then setting $r_k(x):=c_{-1}x^{-1}+\sum_{j=1}^{n_k}c_j x^{n+m_j}$ then the $r_k$ are rational functions and $r:=\prod_{k=1}^n r_k$ is a rational function with the form $r(x)=\gamma x^{-1}+p(x)$ for some polynomial $p\in F[x]$ such that $p(0)=0$ and $\gamma\neq 0$. Also by construction we find that $r(\tau)=0$ what imply that $\tau^{-1}=-\gamma^{-1} p(\tau)$, what finishes the proof.
EDIT: ¡ah! There is a little flaw in the proof above, I would say that $r(x)=\gamma x^{-n}+p(x)$ instead, then $\tau^{-1}=-\gamma^{-1}\tau^{n-1}p(x)$ holds.
Also the condition $p(x)=0$ was unnecesary, I dont know what I was thinking to impose this condition.
What about the following argument? (This avoids the use of Caley-Hamilton theorem.) Let us consider $$\{\iota,\tau,\ldots,\tau^{n^2+1}\}\subset \mathcal{L}(V).$$ Since $\dim \mathcal{L}(V)=n^2<\infty$, it follows that $\iota,\tau,\ldots,\tau^{n^2+1}$ are linearly dependent and there exists a non-zero polynomial $p(t)\in\mathbb{F}[t]$ such that $p(\tau)=O$. Let $m(t)$ be the monic polynomial of the smallest degree for which $m(\tau)=O$ (i.e. $m$ is the minimal polynomial of $\tau$.) If $m(0)=0$, then we may write $$ m(\tau)=m^*(\tau)\cdot\tau=O $$ where $m^*(t)=\frac{m(t)}{t}\neq 0$ is a polynomial of degree $\deg m -1$. This impiles $m^*(\tau)=O$ and contradicts the minimality of $m$. This shows that $m(0)=\lambda \neq 0$ and that $$ m^{**}(\tau)\cdot\tau = -\lambda \iota $$ for $m^{**}(t)=\frac{m(t)-m(0)}{t}$. It follows that $\sigma = -\frac{1}{\lambda}m^{**}(\tau)$ as desired.