Prove that:
$\displaystyle \lim_{n\rightarrow \infty} \sum_{i = 0}^{n}\left(\sum_{j = 0}^{i} a^j\right)^{r^i} = \lim_{n\rightarrow \infty} \prod_{j = 0}^{1}\sum_{i = 0}^{n} (a^{j}\cdot r)^i$
Prove that:
$\displaystyle \lim_{n\rightarrow \infty} \sum_{i = 0}^{n}\left(\sum_{j = 0}^{i} a^j\right)^{r^i} = \lim_{n\rightarrow \infty} \prod_{j = 0}^{1}\sum_{i = 0}^{n} (a^{j}\cdot r)^i$
to reverse summation order we only play with inqualities. (this trick is also useful when reversing integration order on double integrals.)
thus, for both sums, we have $\displaystyle 0\le i\le n$ & $\displaystyle 0\le j\le i,$ now just combine these inequalities to get $\displaystyle 0\le j\le i\le n,$ and finally split this multiple inequality into two ones to get the desired order of summation.