Most often a contact manifold is given as a pair $(M,\xi)$ where $\xi$ is a completely non-integrable hyperplane field on $M$---that is, $\xi$ is given (locally) as the kernel of a 1-form $\alpha$ such that $\alpha\wedge(d\alpha)^n\neq 0$.
However, there is an alternative definition that I have never quite understood, that $(d\alpha)^n\vert_\xi\neq 0$. I don't quite know why this is equivalent to $\alpha\wedge(d\alpha)^n\neq 0$.
Any hints would be appreciated!
Computational approach. If you don't mind I will do the computations for $n=1$.
Let $X_1,X_2,X_3$ be vector fields on $M$, one has the following equality: $$2(\alpha\wedge\mathrm{d}\alpha)(X_1,X_2,X_3)=\sum_{\sigma\in\mathfrak{S}_3}\varepsilon(\sigma)\alpha(X_{\sigma(1)})\mathrm{d}\alpha(X_{\sigma(2)},X_{\sigma(3)})\tag{1}.$$
Going back to the result we aim for.
Let $x\in M$, there exists $U$ a neighborhood of $x$ and $X,Y,Z\in\Gamma(TU)$ such that: $$\ker(\alpha)_{\vert U}=\textrm{Span}(X,Y),TU=\ker(\alpha)_{\vert U}\oplus\mathbb{R}Z.$$
Assume $\alpha\wedge\mathrm{d}\alpha=0$, then using $(1)$, one gets the following equality: $$\alpha(Z)\mathrm{d}\alpha(X,Y)=0.$$ Therefore, $\mathrm{d}\alpha(X,Y)=0$ and by linearity $\mathrm{d}\alpha_{\vert U}$ vanishes identically on $\ker(\alpha)_{\vert U}$.
Assume that $\mathrm{d}\alpha$ vanishes identically on $\ker(\alpha)$, then using $(1)$, one gets the following equality: $$\alpha\wedge\mathrm{d}\alpha(X,Y,Z)=0.$$ Therefore, by linearity $\alpha\wedge\mathrm{d}\alpha_{\vert U}$ vanishes identically on $TU$.
Whence the result taking the contrapositive.
In summary, we have established that $\alpha$ is a contact form if and only if $\mathrm{d}\alpha$ is a symplectic form on $\ker(\alpha)$.
Linear algebra approach. Let us prove the following general result:
Proof. First, assume that $\omega=\alpha\wedge\beta$, then $\alpha\wedge\omega=0$.
Assume that $\alpha\wedge\omega=0$, let $\alpha_1:=\alpha\neq 0$ and let us complete $(\alpha_1)$ as a basis $(\alpha_1,\ldots,\alpha_n)$ of $E^*$. Therefore, one can write $\omega$ in a unique way as: $$\omega=\sum_{1\leqslant i_1<\ldots<i_k\leqslant n}\omega_{i_1,\ldots,i_k}\alpha_1\wedge\cdots\wedge\alpha_k,$$ from our assumption, one gets: $$\sum_{1<i_1<i_2<\ldots<i_k\leqslant n}w_{i_1,\ldots,i_k}\alpha\wedge\alpha_{i_1}\wedge\cdots\wedge\alpha_{i_k}=0,$$ but $\{\alpha\wedge\alpha_{i_1}\wedge\cdots\wedge\alpha_{i_k}\}_{1<i_1<i_2<\ldots<i_k\leqslant n}$ is linearly independent, so that if $1\not\in\{i_1,\ldots,i_k\}$, $\omega_{i_1,\ldots,i_k}=0$. Finally, one gets: $$\omega=\sum_{1<i_2<\ldots<i_k\leqslant n}\omega_{1,i_2,\ldots,i_k}\alpha\wedge\alpha_{i_2}\wedge\cdots\wedge\alpha_{i_k}=\alpha\wedge\beta.$$
Assume that $\omega=\alpha\wedge\beta$, then $\omega_{\ker(\alpha)}=0$ using that: $$\alpha\wedge\beta(x_1,\ldots,x_k)=\frac{1}{(k-1)!}\sum_{\sigma\in\mathfrak{S}_k}\varepsilon(\sigma)\alpha(x_{\sigma(1)})\beta(x_{\sigma(2)},\ldots,x_{\sigma(k)}).$$
Assume that $\omega_{\vert\ker(\alpha)}=0$, then $\alpha\wedge\omega=0$, using that: $$\alpha\wedge\omega(x_1,\ldots,x_{k+1})=\frac{1}{k!}\sum_{\sigma\in\mathfrak{S}_{k+1}}\varepsilon(\sigma)\alpha(x_{\sigma(1)})\omega(x_{\sigma(2)},\ldots,x_{\sigma(k+1)}),$$ along with the splitting $E=\ker(\alpha)\oplus\langle R_{\alpha}\rangle$.
Whence the result. $\Box$
Whence the result, working pointwise in the tangent bundle and taking the contrapositive of the proposition.
Addendum. If you are also wondering why $\ker(\alpha)$ is integrable if and only if $\alpha\wedge\mathrm{d}\alpha=0$, this comes from: $$\mathrm{d}\alpha(X,Y)=X(\alpha(Y))-Y(\alpha(X))-\alpha([X,Y]),$$ which holds for all vector fields on $M$ along with Frobenius theorem.