Do the $x^{th}$ differences of $n^x$ always equal $x!$ Let's take the series $n^3$:

*

*$0^3 = 0$

*$1^3 = 1$

*$2^3 = 8$

*$3^3 = 27$

*etc.

Now, we can take the differences of the series:
0   1   8   27   64   125   216     # Original Series

  ^   ^   ^    ^    ^     ^
  1   7  19   37   61    91         # 1st differences

    ^   ^   ^    ^    ^
    6  12  18   24   30             # 2nd differences

      ^   ^   ^    ^
      6   6   6    6                # 3rd differences

It seems that, for the series $n^x$, the $x^{th}$ differences are all equal to $x!$
However, I don't know how to prove this, and so I cannot be sure if this is true for all positive integer values of $x$.
Does anyone know how to prove/disprove this?
 A: Let us examine the first successive differences of a generic monomial, $ax^n$, without specifying what $a$, $x$, or $n$ we're putting in.
We have $a(x+1)^n - ax^n$; the first half expands via the binomial theorem to $ax^n + nax^{n-1}+\cdots$, so we can subtract out the main term and get the result $nax^{n-1}+\cdots$.  There's more terms, but it will turn out not to matter, as we'll see in a moment.
There are two key observations here:

*

*the result of successive differences on this monomial has reduced its degree by $1$, and if the degree was already $0$ (a constant), now the whole monomial has gone to $0$.

*it also multiplies the highest degree term by $n$.

So if we do it twice (assuming $n$ is high enough) then we have reduced the degree of the highest term by $2$ and multiplied it by $n(n-1)$.  And if we do it $k$ times, we have reduced the degree of the highest term by $k$ and multiplied it by $n(n-1)\cdots(n-k+2)(n-k+1)$.
That's looking awfully familiar.
What if we do it $n$ times?  The degree is now $n-n = 0$ and the coefficient has been multiplied by $n(n-1)\cdots2\cdot1$... and that's the factorial $n!$, so we're pretty much done.
And those lower-degree terms?  Well, we had (for instance) $x^{n-2}$ after one application of successive differences, but we have to do successive differences $n-1$ more times, so before it gets all the way through, it had degree $0$, which means it fell to $0$ by the time we'd finished the job.
A: COMMENT.-In general, if we give a polynomial $f(x)=a_0x^n+\cdots +a_n$ values in arithmetic progression of ratio $h$, the difference of order $n$ of said polynomial is constant. It is easy to check that the difference of order $n-m$ is a polynomial of degree $n-m$ whose first term is $n(n-1)...(n-m+1)a_ox^{n-m}h^n$. It follows that the difference of order $n$ will be $$\Delta^nf(x)=n!a_0h^n$$ This is internally tied to the derivative of order $n$ of the polynomial whose expression we know.
In particular we have obviously what you feel about your monomial.
