wedge product of differential form If $\alpha $ is one form over some manifold $M$ $2n-1$ dimensional real, and $X= M\times (0,\infty)$. $r$ is the coordinate for the second factor.   Define two form on $X$: 
$$\omega= d(r^2\alpha)$$
Then we have to calculate $\omega^n$.   I am sorry if following doubt are too silly: My doubts are:
1-  I think, $\omega^n:= \omega\wedge..\wedge\omega$, n times.
2-  As $\omega$ is two form hence $\omega\wedge \omega \neq 0$.  but for any one form $\alpha$, we must have $\alpha\wedge\alpha= 0$.[As $\alpha\wedge\alpha= c(\alpha\otimes \alpha- \alpha\otimes \alpha)$
3- What is the guarantee that $\omega^n\neq 0$, As in one form(as in 2nd part above), can we say when $\omega^n=0$ for any two form.
 A: *

*That depends on the convention in your textbook or notes. But I have seen that notation used, so I won't rule it out.


*"As $\omega$ is a two-form hence $\omega\wedge\omega\neq 0$" is false. Consider the two form $\mathrm{d}x\wedge \mathrm{d}y$ on $\mathbb{R}^4 = \{(w,x,y,z)| w,x,y,z\in\mathbb{R}\}$. This two form wedged with itself is zero. What you meant is that "there exists two forms such that $\omega\wedge \omega \neq 0$", which is true.


*There are some cases when $\omega^n = 0$ can be easily guaranteed by algebraic constraints. For starters, if the dimension of $X$ is less than $2n$, then since $\omega^n$ is a $2n$-form, it must be identically 0. This can be generalised using the rank of the differential form using the notion of form envelope.
Let $V$ be a vector space and let $\eta \in \wedge^p V$. We can consider the smallest subspace $W\subseteq V$ such that $\eta \in \wedge^p W$. It is clear that if $kp > \mathrm{dim}(W)$ that $\eta^k = 0$. It is easy to see that since every one form lives in a one dimensional sub-space, this means that for one-forms $\alpha^k = 0$ if $k > 1$.
For your specific form, you have that
$$ \omega = 2 r\mathrm{d}r \wedge \alpha + r^2 \mathrm{d}\alpha $$
The main constraint then is what $\mathrm{d}\alpha$ looks like. But noting that
$$ (\mathrm{d}r \wedge \alpha)^2 = 0 $$
you can directly compute (using basically the binomial formula) that
$$ \omega^n = r^{2n} (\mathrm{d}\alpha)^n + 2n r^{2n-1} \mathrm{d} r \wedge \alpha \wedge (\mathrm{d}\alpha)^{n-1} ~.$$
Whether this vanish is determined by whether $(\mathrm{d}\alpha)^n$ vanishes and whether $\alpha\wedge(\mathrm{d}\alpha)^{n-1}$ vanishes.
Now, since per your edit, $M$ is $2n-1$ dimensional. Necessarily $(\mathrm{d}\alpha)^n$ which is a $2n$ form on $M$ must be zero. So you are reduced to
$$ \omega^n = 2n r^{2n-1} \mathrm{d}r\wedge\alpha\wedge(\mathrm{d}\alpha)^{n-1} $$
This is the best you can do in the abstract, unless more information about $\alpha$ is given. As an example, take $n = 2$ and $M = \mathbb{R}^3$ with coordinates $x,y,z$. Let $\alpha = \mathrm{d}z + x\mathrm{d}y$. Then $\mathrm{d}\alpha = \mathrm{d}x\wedge \mathrm{d}y$ and
$$ \omega^2 = 4 r^3 \mathrm{d}r \wedge \mathrm{d}z \wedge \mathrm{d}x\wedge \mathrm{d}y~.$$
