Since $a$ is an even positive integer, it follows that $A$ is odd. Therefore $A$ will be a perfect square of an odd integer, that is
$$A = (2\kappa + 1)^2 = 4\kappa^2 + 4\kappa + 1 = 4\kappa(\kappa + 1) + 1,$$
where $\kappa$ is a positive integer. Since one of the two integers $\kappa$ and $\kappa + 1$ is even, we have
$$\begin{align*} A &= 4\kappa(\kappa + 1) + 1 = 8\rho + 1, \text{ where } \rho \text{ is a positive integer} \\ \Rightarrow A - 1 &= a^n + a^{n-1} + \dots + a = 8\rho \Rightarrow a(a^{n-1} + \dots + a + 1) = 8\rho \\ \Rightarrow 8|a(a^{n-1} + \dots + a + 1) \Rightarrow 8|a, \text{ since } (8, a^{n-1} + \dots + a + 1) = 1. \end{align*}$$