$\sum_{k=0}^n \sum_{j=0}^k { {n} \choose {j}} { {n-j} \choose {k-j}} \left( \frac{x}{1-x}\right)^k = \left( \frac{1+x}{1-x} \right)^n$ How to prove the following formula?
$$\sum_{k=0}^n \sum_{j=0}^k { {n} \choose {j}} { {n-j} \choose {k-j}} \left( \frac{x}{1-x}\right)^k = \left( \frac{1+x}{1-x} \right)^n$$
I have tried to write $(1+x)^n$ with binomial formula in the upstairs but I don't see how to include the downstairs into this. This formula actually arises from enumerating supports of squarefree monomials of $\mathbb{F}[x_1,\dots, x_n, y_1, \dots, y_n]$ written with multi-indeces as $x^a y^b$ such that $a \cdot b = 0$. I think it can be done with considering the simplicial complex they form as a iterated join of one that is just two points. But I'd like to prove the formula in a more "simple" way.

Let $ n $ be a positive integer.
\begin{aligned} \sum_{k=0}^{n}{\sum_{j=0}^{k}{\binom{n}{j}\binom{n-j}{k-j}\left(\frac{x}{1-x}\right)^{k}}}&=\sum_{j=0}^{n}{\sum_{k=j}^{n}{\binom{n}{j}\binom{n-j}{k-j}\left(\frac{x}{1-x}\right)^{k}}}\\ &=\sum_{j=0}^{n}{\binom{n}{j}\left(\frac{x}{1-x}\right)^{j}\sum_{k=0}^{n-j}{\binom{n-j}{k}\left(\frac{x}{1-x}\right)^{k}}}\\ &=\sum_{j=0}^{n}{\binom{n}{j}\left(\frac{x}{1-x}\right)^{j}\left(1+\frac{x}{1-x}\right)^{n-j}}\\ &=\left(\frac{x}{1-x}+1+\frac{x}{1-x}\right)^{n}\\ \sum_{k=0}^{n}{\sum_{j=0}^{k}{\binom{n}{j}\binom{n-j}{k-j}\left(\frac{x}{1-x}\right)^{k}}}&=\left(\frac{1+x}{1-x}\right)^{n}\end{aligned}
