Why is there a pattern in exponents where x^y and (x+1)^y increases by y! "accelerated" by y in? I've noticed with exponents a certain pattern that occurs.
1^2=1 | 2^2=4 | 3^2=9 | 4^2=16 | 5^2=25
1+3 =4 | 4+5 =9 | 9+7=16 | 16+9=25
You find 3, 5, 7, and 9; all 2 in between.
It takes 2 times of taking the difference to reach a constant number.
1^3=1 | 2^3=8 | 3^3=27 | 4^3=64 | 5^3=125
1+7=8 | 8+19=27 | 27+37=64 | 64+61=125
7+12=19 | 19+18=37 | 37+24=61
12, 18, and 24 are all 6 apart; and it takes three (since it cubed) times to get there.
I've tried this more times, I left it out so it doesn't take too much space.
It comes to a constant 2 (or 2!) after 2 times for squared, and 6 (or 3!) with it taking three times to get there.
It makes sense that there would be a pattern like that, but I can't understand why or find any proof.
 A: Let's frame your claim in the language of finite differences, defined as
$$\Delta f(x):=f(1+x)-f(x).$$
We define iterated differencing recursively for positive integer $n$:
$$\Delta^nf(x)=\Delta (\Delta^{n-1}f(x)),\quad \Delta^0f(x)=f(x).$$
You want to show for any integer $n\geq 0$,
$$\Delta^n x^n=n!\quad (1)$$
This claim is quite analogous to
$$\frac{d^n}{dx^n}x^n=n!$$
and both claims can be shown by induction.

Let's show $(1)$ by induction. Trivially, $(1)$ holds for the base case $n=0$.  Assume $(1)$ holds for $n=N-1$; we will show it holds for $n=N.$
By the Leibniz rule, we have in general,
$$\Delta^n f(x)g(x)=\sum_{k=0}^n {n \choose k}(\Delta^kf(x))(\Delta^{n-k}g(x+k)).\quad (2)$$
Apply $(2)$ to your setup by taking $f(x)=x,g(x)=x^{N-1}$. Note $\Delta^k x=0$ for $k\geq 2$, so we only need to compute the first two terms in the sum of $(2)$:
$$\Delta^{N}x^{N}=\Delta^{N}xx^{N-1}={N \choose 0}(\Delta^0 x)(\Delta^{N}x^{N-1})+{N \choose 1}(\Delta^1 x)(\Delta^{N-1}(1+x)^{N-1}).\quad (3)$$
Note by the inductive hypothesis, the first term in $(3)$ is zero since
$$(\Delta^{N}x^{N-1})=\Delta(\Delta^{N-1}x^{N-1})=\Delta (N-1)!=0.$$
Now look at the second term of $(3).$ Since $x$ is arbitrary, our inductive hypothesis implies $$\Delta^{N-1} (x+1)^{N-1}=(N-1)!$$ Thus,
$$\Delta^N x^N={N \choose 1}(\Delta^1 x)(\Delta^{N-1}(1+x)^{N-1})=N(1)(N-1)!=N!$$
and we are done.
