I found a correlation between n^x and x! can anyone explain it? This is the example to numbers to the 3rd.
Here is the same thing just with numbers to the 4th
Well I was messing around with numbers I made this discovery, with the correlation between n^x and x!. What is the reason for this? Is it easily explainable?
In the sheets every column is found from the differences of two numbers in sequence in the last column
 A: The relevant keyword here is finite differences. In general, if $f(n)$ is a sequence, we can construct from it a new sequence, the forward difference
$$(\Delta f)(n) = f(n+1) - f(n).$$
There is also a backward difference. For example, if $f(n) = n^2$, then
$$(\Delta f)(n) = (n + 1)^2 - n^2 = 2n + 1.$$

Exercise #1: Show that if $f(n) = an^d + \dots $ is a polynomial of degree $d$ with leading coefficient $a$, then $(\Delta f)(n) = d an^{d-1} + \dots$ is a polynomial of degree $d - 1$ with leading coefficient $da$.
Exercise #2: Use induction on exercise #1 to show that if $f(n) = n^d$, then $\Delta^d f$ (the result of taking the forward difference $d$ times) is the constant sequence with constant value $d!$.

If you've taken calculus this should remind you a lot of what happens when you differentiate polynomials. There are several more very nice things to say about finite differences but this is enough to explain your observation.
A: The binomial theorem says
$$
(a+b)^x = a^x + xa^{x-1}b + \frac{x(x-1)}2 a^{x-2} b^2 + \frac{x(x-1)(x-2)} 6 a^{x-3} b^3 + \cdots.
$$
The differences in the column to the right of the one that lists values of $a^x$ are
$$
(a+1)^x - a^x = \left( a^x + x a^{x-1} + \frac{x(x-1)} 2a^{x-2} + \cdots \right) - a^x.
$$
Since $a^x$ cancels, you have $xa^{x-1}$ as the highest-degree term in the resulting polynomial function of $a.$
Then for the next row, for the same reason you get $x(x-1)x^{x-2}$ as the highest-degree term.
And next you get $x(x-1)(x-2)a^{x-3},$ and so on until finally you have $x(x-1)(x-2)\cdots 1 a^{x-x}.$
