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I read that the $n$-th finite difference of the sequence

$1^n, 2^n, 3^n,\dots$

is $n!$, but I'm not able to prove this. Could someone give an idea of why this is true?

Tian An
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    Since you just want an idea of why this is true, why don't you try computing the first differences of (1, 2, 3, 4, 5, ...), the second differences of (1, 4, 9, 16, 25, ...), the third differences of (1, 8, 27, 64, 125, ...), and see what happens? – Unit Apr 21 '16 at 00:55
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    Hint: look at the first difference. What's the lead term? That's the only term that matters... If you now try to take the $(n-1)^{st}$ difference of the first difference, all the terms of order $â‰Īn-2$ will vanish. – lulu Apr 21 '16 at 01:00
  • The hints were good at giving the intuition. Thanks! – Tian An Apr 21 '16 at 01:54

1 Answers1

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The difference operator $\delta$ that maps a polynomial $p(x)$ into $(\delta p)(x)=p(x+1)-p(x)$ has the following properties:

  • If the degree of $p(x)$ is $d\geq 1$, the degree of $\delta p$ is $d-1$;
  • If the leading term of $p(x)$ is $c_d x^d$, the leading term of $(\delta p)(x)$ is $d c_d x^{d-1}$ (the same as $p'(x)$).

If follows that if $p(x)=x^n$ or $(x+1)^n$, $\delta^n p$ is a polynomial with degree zero (i.e. a constant) and leading term $n\cdot(n-1)\cdot\ldots\cdot 1 = \color{red}{n!}$.

An alternative proof. We want to show that: $$ \sum_{k=0}^{n}\binom{n}{k}(-1)^{n-k} k^n = n!. $$ $k^n$ is the number of functions from $\{1,2,\ldots,n\}$ to $\{1,2,\ldots,k\}$, hence the LHS accounts for the functions $f:\{1,2,\ldots,n\}\to \{1,2,\ldots,n\}$ such that $|\text{Im}\,f|=n$, i.e. the surjective functions, that clearly are $\color{red}{n!}$.

A third alternative. Let: $$ A(n) = \sum_{k=0}^{n}\binom{n}{k}(-1)^{n-k} k^n. $$ Since $k\binom{n}{k} = n\binom{n-1}{k-1}$, we have: $$ A(n) = n\sum_{k=1}^{n}\binom{n-1}{k-1}(-1)^{n-k} k^{n-1} = n\sum_{k=0}^{n-1}\binom{n-1}{k}(-1)^{n-1-k} k^{n-1} = n\cdot A(n-1).$$ Since $A(1)=1$, $A(n)=\color{red}{n!}$ follows by induction.

Jack D'Aurizio
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