Logarithms equality

Let us start here:

$$\log_b(a)\log_b(n)=\log_b(n)\log_b(a)$$

$$b^{\left(\log_b(a)\log_b(n)=\log_b(n)\log_b(a)\right)}$$

$$b^{\log_b(a)\log_b(n)}=b^{\log_b(n)\log_b(a)}$$

$$\left(b^{\log_b(a)}\right)^{\log_b(n)}=\left(b^{\log_b(n)}\right)^{\log_b(a)}$$

Recall that $x^{\log_x(y)}=y$, so

$$a^{\log_b(n)}=n^{\log_b(a)}$$

For this particular nut, since $n =b^i$ your equality is just $$a^{\displaystyle \log_b b^i} = a^{\displaystyle i} = b^{\displaystyle \log_b a^i}$$

and so it holds. It's also true for a more general $n$, as the comments explain.

It's true for all $a,b,n>0$

\begin{align*} \large \frac{\log a}{\log b} \times \log n &= \large \frac{\log n}{\log b} \times \log a \\[5pt] \large \log_{b} a \times \log n &= \large \log_{b} n \times \log a \\[5pt] \large \log \left( n^{\log_{b} a} \right) &= \large \log \left( a^{\log_{b} n} \right) \\[5pt] \large n^{\log_{b} a} &= \large a^{\log_{b} n} \end{align*}