Geometric series as a power series

Remarks

The proof is a telescoping product. The coefficient of $X^n$ in the product is $a^n - a \cdot a^{n-1} = 0$ for $n \ge 1$, and it is $1$ for $n = 0$. With $a = 1$, it gives $\frac{1}{1 - X} = 1 + X + X^2 + \dots$, the generating function of the constant sequence $1$. Over $\mathbb{R}$ the same identity holds as a convergent series for $|aX| < 1$, but the formal identity needs no convergence.