Suppose $ST=1_V$ and $S, T \in Hom_F(V, V)$, where V is a finite dimensional vector space over F. Show $S=f(T)$ I want to show that there is a polynomial $f\in F[x]$ such that $S=f(T)$ but I don't know what properties to look for.  
I would like to show this without Cayley-Hamilton theorem if possible

Since $V$ is finite-dimensional, so is $\operatorname{Hom}(V,V)$.  It follows that there must be some linear relation between the infinitely many elements $1_V, T, T^2, T^3, \dots,\in \operatorname{Hom}(V,V)$.  That is, there is some nonzero polynomial $p(x)\in F[x]$ such that $p(T)=0$.  Now write $p(x)=x^nq(x)$ where $q(x)$ has a nonzero constant term and $n\geq 0$.  Then $T^n q(T)=0$, and multipling both sides by $S^n$ on the left gives $q(T)=0$.  That is, $\sum_{i=0}^m a_i T^i=0$ for some scalars $a_0,\dots,a_m$ with $a_0\neq 0$.  We then find that $$T\left(-\sum_{i=1}^m\frac{a_i}{a_0} T^{i-1}\right)=1_V.$$
That is, taking $f(x)=-\sum \frac{a_i}{a_0}x^{i-1}$, we have $Tf(T)=1_V$.  Multiplying both sides by $S$ on the left, we now get $f(T)=S$.
