Use List::Util::sum for summation: 
#! /usr/bin/perl
use warnings;
use strict;
use feature qw{ say };

use List::Util qw{ sum };
for my $A (2..20) {
    for my $n (1..10) {
        die "$A, $n\n"
            unless $A ** $n == 1 + ($A - 1) * sum(map $A ** $_, 0 .. $
+n - 1)
    }
}

[download]

The proof can be compiled in TeX: 

\documentclass{article}
\title{Summation Fulguration}
\newtheorem{thm}{Theorem}
\newtheorem{prf}{Proof}[thm]
\begin{document}
\begin{thm}
$$a^n = 1 + (a - 1) \sum_{i=0}^{n-1} a^i$$
\end{thm}
\begin{prf}
$$1 + (a - 1) \sum_{i=0}^{n-1} a^i$$
$$= 1 + \sum_{i=0}^{n-1}(a-1)a^i$$
$$= 1 + \sum_{i=0}^{n-1} a^{i+1} - a^i$$
$$= 1 + \sum_{i=0}^{n-1}a^{i+1} - \sum_{i=0}^{n-1}a^i$$
$$= 1 + (\sum_{i=1}^{n-1}a^i) + a^n - (a^0 + \sum_{i=1}^{n-1}a^i)$$
$$= 1 + (\sum_{i=1}^{n-1}a^i) + a^n - a^0 - \sum_{i=1}^{n-1}a^i$$
$$= 1 + a^n - a^0$$
$$= 1 + a^n - 1$$
$$= a^n$$
Q.E.D.
\end{prf}
\end{document}

[download]

The result can be checked here.

I have no idea how broadly it is known.

map{substr$_->[0],$_->[1]||0,1}[\*||{},3],[[]],[ref qr-1,-,-1],[{}],[sub{}^*ARGV,3]

In reply to Re: [OT] math fulguration by choroba
in thread [OT] math fulguration by Discipulus

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