So in this proof (please don't ask about it, it's not important and it would take ages to explain) there's this step where they "switch" the values of the series of the double products in the following way.
Say you have a function with $2$ variables, $f(i,j)$ and the first product goes from $i=1$ to $i=j$ and the 2nd product goes from $j=1$ to $j=k$, where $k$ is a constant:
$$\prod_{j=1}^{k} \prod_{i=1}^{j} f(i,j)$$
and the proof "switches the order" saying that the above is equivalent to
$$\prod_{i=1}^{k} \prod_{j=i}^{k} f(i,j)$$
So basically an explicit explanation of how the two equations are equivalent would be great. I have tried expanding it out but i end up getting confused so hearing as many explanations as possible would really help me out.
After the latest comment by the OP, I edited again to fix the typo in the indexing of the final product. The two products are indeed equivalent. To see this, I recommend drawing a picture, with $i$ on the horizontal axis and $j$ on the vertical axis. In both cases, the product is taken over all $(i,j)$ coordinates on and above the diagonal.
In the first case, $$\prod_{j=1}^{k} \prod_{i=1}^{j} f(i,j)$$ the product is first taken along each row (to the left of and including the diagonal), and then the row products are multiplied together.
In the second case, $$\prod_{i=1}^{k} \prod_{j=i}^{k} f(i,j)$$ the product is first taken along each column (above and including the diagonal), and then the column products are multiplied together.
Both products are equal to $$\prod_{\substack{i,j = 1\\i \leq j}}^{k} f(i,j)$$