<s>[INST] 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.
 [/INST] What you have discovered is a finite difference calculation. For any function $f$, in this case the third-power function
$$
f(n) = n^3
$$
we can define the forward difference, or forward discrete derivative:
$$
\Delta f(n) = f(n+1) - f(n) = 3n^2 + 3n + 1
$$
Likewise,
\begin{align*}
\Delta \Delta f(n) = \Delta^2 f(n) &= 6n+ 6 \\
\Delta^3 f(n) &= 6 \\
\Delta^4 f(n) &= 0.
\end{align*}
Your computation,
$$
5^3 = 64 + 37 + 18 + 6
$$
is the statement
$$
f(5) = f(4) + \Delta f(3) + \Delta^2 f(2) + \Delta^3 f(1),
$$
or more generally
$$
f(n) = f(n-1) + \Delta f(n-2) + \Delta^2 f(n-3) + \Delta^3 f(n-4).
$$
This is one discrete analogue of Taylor series
(the more common analogue is Newton's series).
The reason it works is that, for $f(n) = n^3$, $\Delta^4$ and beyond are all zero. So the summation stops once we get to $\Delta^3$.

EDIT:
Let me add that there is yet another identity here which resembles Taylor series, namely,
$$
f(n-1) = f(n) - \Delta f(n) + \Delta^2 f(n) - \Delta^3 f(n).
$$
Altogether, at least when $f$ is a polynomial, we thus have the following Taylor series-like formulas:
\begin{align*}
f(n-1) &= \sum_{i = 0}^\infty (-1)^i [\Delta^i f](n) \\
f(n+1) &= \sum_{i = 0}^\infty [\Delta^i f](n - i) \\
f(n+x) &= \sum_{i=0}^\infty \binom{x}{i} [\Delta^i f](n).
\end{align*}
</s>