Suppose that $M$ is a smooth $n-$manifold. Suppose $$ TM=\coprod_{p\in M}T_pM $$ is the tangent bundle of $M$. And let $$ \Lambda^n(M)=\coprod_{p\in M}\Lambda^n(T_pM) $$ where $\Lambda^n(T_pM)$ is alternating covariant $n-$tensor product. i.e.
$$\Lambda^n(T_pM)=T^*_pM\wedge T^*_pM \wedge \cdots \wedge T^*_pM.$$
and
$$dim(\Lambda^n(T_pM))=\begin{bmatrix} dim(T_pM)\\ n \end{bmatrix}$$
If $TM$ is trivial, then $\Lambda^n(M)$ is also trivial and $M$ is orientable.
I am trying to understand but seems difficult. I would be very appreciate any help. Thanks in advance.
If $M$ is $n$-dimensional, $TM$ is trivial is equivalent to saying that there exists $n$-vector fields $X_1,...,X_n$ such that for every $x\in M$, $X_i(x)\neq 0$. Take a differentiable metric on $M$ and define the $1$-form $f_i(x)(u)=\langle X_i(x),u\rangle$ where $u\in T_xM$, $\Lambda^n(M)_x$ is generated by $f_1(x)\wedge...\wedge f_n(x)$ which is also a volume form on $M$. This implies that $M$ is orientable.