Is $\left(\sum_{i=0}^{n-1} a^{2i}\right)\left(\sum_{i=0}^{n-1} a^{2i}\right)=\sum_{i=0}^{n-1} a^{4i}$ by distributivity?

You're applying the distributive property like $$ (a+b)(c+d)=ac+bd $$ which is incorrect.

The left-hand side is, for $a\ne\pm1$, $$ \dfrac{a^{2n}-1}{a^2-1}\dfrac{a^{2n}-1}{a^2-1} $$ and the right-hand side is $$ \dfrac{a^{4n}-1}{a^4-1} $$ They could be equal if and only if $$ \frac{a^{2n}-1}{a^2-1}=\frac{a^{2n}+1}{a^2+1} $$ that is $$ a^{2n+2}-a^2+a^{2n}-1=a^{2n+2}+a^2-a^{2n}-1 $$ so when $a^{2n}=a^2$. Since we assumed $a\ne\pm1$, only $a=0$ remains if $n>1$.

If $a=\pm1$, the left-hand side is $n^2$ and the right-hand side is $n$, so equality holds only for $n=1$ (or $n=0$, if you allow empty sums).

Counterexample: $(1+a^2)^2 = 1+2a^2+a^4$.