I started to read about contact and symplectic forms and I came across this answer here.
It seems to state that the definition of symplectic form is that $d\alpha$ is non-degenerate if and only if $(d\alpha )^n \neq 0$ (pointwise).
Everywhere else non-degeneracy of a $2$-form $\beta$ is given as $\beta$ is non-degenerate if and only if $\beta(v,w)=0$ for all $w \in V$ implies $v =0$. Here $V$ is the vector space on which $\beta$ acts.
But it's not clear to me why these two definitions of non-degeneracy are equivalent.
Please could someone explain to me why $(d\alpha )^n \neq 0$ if and only if $d\alpha(v,w)=0$ for all $w \in V$ implies $v =0$?
they're not!
typically you have a two-form $\omega$ such that $\omega^n$ is non-zero everywhere. The two-form is closed i.e. $d\omega=0.$
The non-zero everywhere is the same as saying that the two-form is non-degenerate everywhere.
It is often the case that there is a one-form such that $$ \omega = d\alpha $$ that is it is exact (which is not always the case) but this is not always the case.
For cotangent bundles, $\omega = d\alpha.$