Computing the wedge product of some differentials I am trying to compute the wedge product of
\begin{align*}
&2x_p \mathrm{d}x_p \bigwedge_{i=1,\cdots,p-1}2(x_i \mathrm{d}x_i
+y_i\mathrm{d}y_i)\bigwedge_{i=1,\cdots,p-1}x_{i+1}\mathrm{d}y_i+y_i\mathrm{d}x_{i+1},\qquad\qquad
\end{align*}
I tried computing first
\begin{align}
2(x_i \mathrm{d}x_i+y_i\mathrm{d}y_i)&\wedge x_{i+1}\mathrm{d}y_i+y_i\mathrm{d}x_{i+1}
\\&=(2x_ix_{i+1})\mathrm{d}x_i\wedge \mathrm{d}y_i+
(2x_iy_{i})\mathrm{d}x_i\wedge \mathrm{d}x_{i+1}
+(2y_i^2)\mathrm{d}y_i\wedge \mathrm{d}x_{i+1}
\\
&=(2x_ix_{i+1})\mathrm{d}x_i\wedge \mathrm{d}y_i+
(2x_iy_{i})\mathrm{d}x_i\wedge \mathrm{d}x_{i+1}
-(2y_i^2)\mathrm{d}x_{i+1}\wedge \mathrm{d}y_i
\end{align}
But here I don't know what to do.
 A: Look like you (or your reference) like to order index in decreasing order, which makes the induction argument a bit weird. I assume $\mathrm{d} \vec x = \mathrm{d} x_p \wedge \mathrm{d} x_{p-1}\wedge \cdots \wedge \mathrm{d} x_1$ and similar for $\mathrm{d}\vec y$.
We prove by induction on $p$.
When $p=2$,
\begin{align}
& (2x_2 \mathrm{d}x_2)\wedge (2(x_1 \mathrm{d}x_1
+y_1\mathrm{d}y_1)) \wedge (x_{2}\mathrm{d}y_1+y_1\mathrm{d}x_{2})\\
&= 2^2 x_2 \mathrm{d}x_2 \wedge (x_1 x_2\mathrm{d}x_1 \wedge\mathrm{d}y_1 + x_1 y_1 \mathrm{d}x_1 + y_1^2 \mathrm{d} y_1\wedge \mathrm{d} x_2) \\
&= 2^2 ( x_1x_2^2 \mathrm{d} x_2 \wedge \mathrm{d}x_1 \wedge\mathrm{d}y_1)  \\
&= 2^2 x_1 x_2^2\mathrm{d} \vec x \wedge\mathrm{d}\vec y. 
\end{align}
So it's true for $p=2$.
Now assume that it's true for $p = p_0$. For $p = p_0+1$,
\begin{align}
A_{p_0+1} &= (2x_{p_0+1} \mathrm{d} x_{p_0+1} ) \wedge \bigwedge_{i=p_0}^1 2(x_i \mathrm{d}x_i
+y_i\mathrm{d}y_i) \wedge \bigwedge _{i=p_0}^1 (x_{i+1}\mathrm{d}y_i+y_i\mathrm{d}x_{i+1}) \\
&= (2x_{p_0+1} \mathrm{d} x_{p_0+1} ) \wedge \bigwedge_{i=p_0}^2 2(x_i \mathrm{d}x_i +y_i\mathrm{d}y_i) \wedge \bigwedge _{i=p_0}^2 (x_{i+1}\mathrm{d}y_i+y_i\mathrm{d}x_{i+1}) \\
& \  \ \  \wedge (-1)^{p_0-1} 2(x_1 \mathrm{d}x_1+y_1\mathrm{d}y_1) \wedge (x_{2}\mathrm{d}y_1+y_1\mathrm{d}x_{2}) 
\end{align}
Now apply the induction hypothesis on
$$ (2x_{p_0+1} \mathrm{d} x_{p_0+1} ) \wedge \bigwedge_{i=p_0}^2 2(x_i \mathrm{d}x_i +y_i\mathrm{d}y_i) \wedge \bigwedge _{i=p_0}^2 (x_{i+1}\mathrm{d}y_i+y_i\mathrm{d}x_{i+1}) $$
to obtain
\begin{align}
A_{p_0+1} &= (2^{p_0}x_2 \prod\limits_{i=3}^{p_0+1} x_i^2)\;\bigwedge_{i=p_0+1}^2 \mathrm{d} x_i\wedge \bigwedge_{i=p_0}^2 \mathrm{d}\vec y\\
& \  \ \  \wedge (-1)^{p_0-1} 2(x_1 \mathrm{d}x_1+y_1\mathrm{d}y_1) \wedge (x_{2}\mathrm{d}y_1+y_1\mathrm{d}x_{2}) \\
&=  (2^{p_0}x_2 \prod\limits_{i=3}^{p_0+1} x_i^2)\;\bigwedge_{i=p_0+1}^2 \mathrm{d} x_i\wedge \bigwedge_{i=p_0}^2 \mathrm{d}\vec y\\
& \  \ \  \wedge (-1)^{p_0-1} 2(x_1x_2 \mathrm{d}x_1\wedge \mathrm{d}y_1 + x_1y_1 \mathrm{d}x_1\wedge \mathrm{d}x_2+y_1^2 \mathrm{d}y_1\wedge \mathrm{d}x_2)\\
&=  (2^{p_0+1}x_1x_2 x_2 \prod\limits_{i=3}^{p_0+1} x_i^2)\;\bigwedge_{i=p_0+1}^2 \mathrm{d} x_i \wedge \mathrm{d}x_1 \wedge \bigwedge_{i=p_0}^2 \mathrm{d}y_i \wedge \mathrm{d}y_1 \\
&= (2^{p_0+1}x_1 \prod\limits_{i=2}^{p_0+1} x_i^2)\;\bigwedge_{i=p_0+1}^1 \mathrm{d} x_i \wedge \bigwedge_{i=p_0}^1 \mathrm{d}y_i. 
\end{align}
Thus it is also true for $p_0+1$. By induction we have proved the assertion.
