Invertible Linear Map Suppose $V$ is finite-dimensional and $S,T\in \mathcal{L}(V)$. Prove that if $ST$ is invertible, then both $S$ and $T$ are invertible. 
This is the partial question retrieved from Linear Algebra Done Right, and I have come out with different solution from the solution guide. 
My solution: 
Since $ST$ is invertible, then the only possibility for $(ST)u=0$ is when $u=0$. Therefore I have $(ST)(0)=0=S(T(0))=0$. Since $T(0)=0$ (as proven in book), I must have $S(0)=0$. Since $T(0)=0$ and $S(0)=0$ comes directly from $(ST)(0)=0$, both $S$ and $T$ are injective. Since both of them are linear operator, we can deduce they are both invertible (by equivalece statement.)
Is my proof valid?

Note that since ST is invertible
$$\forall u\neq 0\iff (ST)u\neq0$$
then since 
$$(ST)u=(S)((T)u)\neq0 \implies v=(T)u\neq0$$
thus T is invertible.
Since T is surjective, we have that $\forall v\neq0 \quad\exists u\neq0 \quad v=(T)u$ and $S(v) \neq 0$, therefore also S is invertible.
