# Property of $\{0^n, 1^n, \ldots\}$ [duplicate]

The differences between adjacent squares are the odd numbers:

0 1 4 9 16 25 …

1 3 5 7 9 …

The differences between adjacent odd numbers are 2 = 2! I found that this truth is more general: If the differences between adjacent powers of $n$ are written out, and the differences of those differences etc, the $(n+1)$th sequence is {n!, n!, …}

Is this a well-known theorem? Does it have an easy proof?

• Yeah, I think this is true; I can't recall a proof at the moment, though. – Akiva Weinberger Sep 18 '15 at 13:34
• Yes. This is the field of "finite differences." – Thomas Andrews Sep 18 '15 at 13:35
• By the way, some notation that you may or may not find useful: Define $\Delta f(x)$ to mean $f(x+1)-f(x)$. Define $\Delta^n f(x)$ to mean $\Delta(\dots(\Delta f(x))\dots)$, with $n\$ $\Delta$s. Then you're asking if $\Delta^n x^n=n!$. – Akiva Weinberger Sep 18 '15 at 13:36
• Related: $(x^n)'=nx^{n-1}$,...,$(x^n)^{(n)}=n!$. – A.Γ. Sep 18 '15 at 13:44
• @user21820, can an older post be a duplicate of a newer? – Toothrot May 8 '19 at 8:25

Take $f(x)$ a polynomial of degree $n$ (in your case, $f(x)=x^2$ or $f(x)=x^k$) and leading coefficient $a$ Then the polynomial $f_1(x)=f(x+1)-f(x)$ has degree $n-1$ and leading coefficient $an$. This is because $$(x+1)^n=x^n+nx^{n-1}+(\text{terms of degree < n-1}).$$ Iterating this procedure (called finite differences) you obtain $f_2(x)$ of degree $n-2$ and leading coefficient $an(n-1)$, $f_3(x)$ of degree $n-3$ and leading coefficient $an(n-1)(n-2)$, etc. After $n$ iterations you get to $f_n(x)$ of degree $n-n=0$ and leading coefficient $an(n-1)\dots 2\cdot 1$, that is $an!$.
• By the way, I think the usual notation uses $\Delta$, as in my comment above. (Correct me if I'm wrong, though.) – Akiva Weinberger Sep 18 '15 at 13:38