Apparently
$dx^{i_1} \wedge ... \wedge dx^{i_k}=d(x^{i_1}dx^{i_2}\wedge ... \wedge dx^{i_k})$
which I cannot see proved anywhere in my notes. It just stated as if it is obvious which I don't believe it is.
I cannot see how you could prove this identity. Firstly, I am not sure as to what $x^{i_1}dx^{i_2}\wedge ... \wedge dx^{i_k}$ is and secondly how you could apply $d$ to it.
On
It is an application of the following fundamental property of the exterior derivative: $d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^{\mathrm{deg} \alpha}(\alpha \wedge d\beta)$.
When $\alpha = f$ is a $0$-form (a function), this identity says that $d(f \beta) = df \wedge \beta + f d\beta$.
In your situation, $f = x^{i_1}$ and $\beta = dx^{i_2}\wedge ... \wedge dx^{i_k}$. You can show that $d \beta = 0$, so you get $d(x^{i_1}\, dx^{i_2}\wedge ... \wedge dx^{i_k}) = dx^{i_1}\wedge dx^{i_2}\wedge ... \wedge dx^{i_k}$.
Oops, I just realized this answer is just what Franck Science wrote in the comments to your question (with a bit more details).
Here is a slightly more systematic answer than in my comment above. Suppose we have the $(k-1)$-form $\omega=f(\vec{x})\omega_{i_1}$ where $\omega_{i_1}:=dx_{i_2}\wedge dx_{i_3}\wedge\cdots \wedge dx_{x_k}$ has been introduced for convenience. Then the exterior derivative of $\omega$ is $$d\omega=df \omega_{x_i}=\left(\frac{\partial f}{\partial x_{i_1}}dx_{i_1}+\frac{\partial f}{\partial x_{i_2}}dx_{i_2}+\cdots + \frac{\partial f}{\partial x_{i_k}}dx_{i_k}\right)\wedge\omega_{i_1}.$$
To simplify this, note that $dx_{i_2}\wedge \omega_{i_1}$=0 since the $1$-form $dx_{i_2}$ is already present in $\omega_{i_1}$. So all the wedge products vanish but the first, yielding
$$d\omega = \frac{\partial f}{\partial x_{i_1}} dx_{i_i}\wedge \omega_{i_1} =\frac{\partial f}{\partial x_{i_1}}dx_{i_1}\wedge dx_{i_2}\wedge dx_{i_3}\wedge\cdots \wedge dx_{x_{i_k}}.$$
For the particular choice of $f(x)=x_{i_1}$ this immediately gives the answer desired since $\dfrac{\partial f}{\partial x_{i_1}}=1$.