Is this a new method for finding powers? Playing with a pencil and paper notebook I noticed the following: 
$x=1$
$x^3=1$

$x=2$
$x^3=8$

$x=3$
$x^3=27$

$x=4$
$x^3=64$

$64-27 = 37$
$27-8 = 19$
$8-1 = 7$

$19-7=12$
$37-19=18$

$18-12=6$

I noticed a pattern for first 1..10 (in the above example I just compute the first 3 exponents) exponent values, where the difference is always 6 for increasing exponentials. So to compute $x^3$ for $x=5$, instead of $5\times 5\times 5$, use $(18+6)+37+64 = 125$.
I doubt I've discovered something new, but is there a name for calculating exponents in this way? Is there a proof that it works for all numbers?
There is a similar less complicated pattern for computing $x^2$ values.

For a little bit more, see the answer "General method for indefinite summation" which explains how exactly this representation using forward differences allows you to easily find the formula for indefinite summation of powers. Applied to your case you get:
0,  1,  8, 27
  1,  7, 19
    6, 12
      6

and hence:

$n^3 = 0 \binom{n}{0} + 1 \binom{n}{1} + 6 \binom{n}{2} + 6 \binom{n}{3}$.

which immediately gives:
$
\newcommand\lfrac[2]{{\large\frac{#1}{#2}}}
$

$\sum_{k=0}^{n-1} k^3 = 0 \binom{n}{1} + 1 \binom{n}{2} + 6 \binom{n}{3} + 6 \binom{n}{4} = n\lfrac{n-1}{2}(1+\lfrac{n-2}{3}(6+\lfrac{n-3}{4}(6)))$


$\ = \lfrac{n^2 (n-1)^2}{4}$.

and then, if you prefer the indices to end at $n$:

$\sum_{k=1}^n k^3 = \lfrac{(n+1)^2 n^2}{4}$.

As you can see, hardly any computation was necessary to get this result!
