$V$ Vector space, Q Skew Symmetric nondgenerate bilnear operator $\rightarrow$ \rightarrow $det(Q)=1$ and dimension of $V$ is even?

Question

Question: If $V$ Vector space, Q Skew Symmetric nondgenerate bilnear operator then show that that $det(Q)=1$ and that dimension of $V$ is even.

My attempt at proof:

Since $Q$ is skew symmetric we have that $$Q^t=-Q$$ So $det(Q)=det(Q^t=det(-Q)=-det(Q)$ edit: WRONG!!! $det(-Q)=(-1)^ndet(Q)$

Hm.. But that's not right.. else $Det(Q)=0$.

Help appreciated!