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.
 A: It's not something new, but for your discovery I applaud. This procedure is called the method of successive differences, and you can show that for every power the successive difference appears.
Let us say you have a sequence:
$$
1^3 \quad2^3\quad 3^3\quad 4^3\quad \ldots
$$
Note that $x^3-(x-1)^3 = 3x^2-3x+1$. So we'll get a new sequence at the bottom:
$$
7 \quad 19\quad 37\quad 61\quad \ldots
$$
Now, note that $3x^2-3x+1-(3(x-1)^2-3(x-1)+1) = 6(x-1)$. Hence, we'll get another series:
$$
0 \quad6\quad 12\quad 18\quad\ldots 
$$
Now, note that $6(x-1)-6((x-1)-1) = 6$!
Now, the new sequence is:
$$
6\quad 6\quad 6\quad 6\quad 6\quad ...
$$
So $6$ appears as the final difference! This shows the power of algebra. As an exercise, do this for $x^4$. See the pattern of the number at the end, and if you can say  something for $x^n$.
The reason, as you can see, is that at each line above, the degree of the polynomial $f(x)-f(x-1)$ decreases by $1$. Hence, at the end of three lines, you are only going to get a constant polynomial.
A: 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*}
A: To test your hypothesis you could work out the form of the differences from the first few cases.
\begin{align*}
1^{3}-0^{3}&=1\\
2^{3}-1^{3}&=7\\
3^{3}-2^{3}&=19\\
4^{3}-3^{3}&=37
\end{align*}
For example rewrite out $(37-19)-(19-7)=18-6=6$ as:
\begin{align*}
\{(4^{3}-3^{3})-(3^{3}-2^{3})\}&-\{(3^{3}-2^{3})-(2^{3}-1^{3})\}\\
&=(4^{3}-2\cdot3^{3}+2^3)-(3^{3}-2\cdot2^{3}+1^{3})\\
&=4^{3}-3\cdot3^{3}+3\cdot2^3-1^{3}\qquad (\star)\\
&=6
\end{align*}
So you have to find the difference of two differences to get to $6$ (this is called a finite difference pattern, and you have to iterate twice to get the result of $6$ for all such differences, any further iteration ending in a $0$). Now check that pattern $(\star)$ holds in general for some integer $k\ge3$:
\begin{align*}
k^{3}&-3\cdot(k-1)^{3}+3\cdot(k-2)^3-(k-3)^{3}\\
&=k^{3}-3(k^2-3k^2+3k-1)
+3(k^3-2\cdot3k^2+2^2\cdot3k-2^3)
-(k^3-3\cdot3k^2+3^2\cdot3k-3^3)\\
&=\ \ k^3\\
&\ -3k^3\ +\ 9k^2\ -\ 9k\ +\  3\\
&\ +3k^3-18k^2+36k-24\\
&\ \ -k^3\ +\ \ 9k^2-27k+27\\
&=6
\end{align*}
A: 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!
