I was a bit hesitant to post this due to the large number of similar questions
on the site, but since I wasn't able to find the details of the second argument
I give anywhere and it is not clear to me which of the other questions should be
considered the canonical duplicate, I am posting it here now.
As I mentioned in my comment, you can prove this using the binomial theorem. The proof is by induction on $n$. If $n = 1$, clearly the differences are all $1 = 1!$. The inductive step goes like this:
Let $n \in \Bbb N$, and consider the sequence $(a + 1)^n - a^n$. By the binomial theorem, this is $\binom n1 a^{n - 1} + \binom n2 a^{n - 2} + \dotsb + 1$. Since $\binom n1 = n$, the $(n - 1)$th-order differences of the first term are $n \cdot (n - 1)! = n!$, by induction. The $(n - 1)$th-order differences of all other terms are $0$, also by induction. The $(n - 1)$th order differences of $(a + 1)^n - a^n$ are therefore $n! + 0 + \dotsb + 0 = n!$. So we are done.
There are many versions of this argument to be found elsewhere on this site (see
these questions:
Problem in solving a question related to finite difference., Is there a simpler way to prove that the nth difference of a degree n polynomial equals n!*(the leading coefficient)?, Elementary proof about nth differences of nth powers of integer, The k-th difference of the sequence $n^{k}$ is constant and equal to $k!$, Repeatedly taking differences on a polynomial yields the factorial of its degree?, How to prove that nth differences of a sequence of nth powers would be a sequence of $n!$, Binomial Summation that Yields Factorial, Why does this process generate the factorial of the exponent?, Property of $\{0^n, 1^n, \ldots\}$, $\Delta^ny = n!$ , difference operator question.)
I've given a somewhat dry, formal version of the argument, but I believe it's
possible to see it as intuitive. Really the key point is that every time you
take the differences of a polynomial, you decrease the degree by $1$ and
multiply the leading coefficient by the degree. Clearly if you do that $n$ times
you'll be left with the constant polynomial $n!$.
In fact to prove that "the differences of a polynomial $P$ of degree $n$ and
leading coefficient $k$ are a polynomial of degree $n - 1$ and leading
coefficient $nk$", you don't need the full binomial theorem - it follows from
the fact that $(t + 1)^n - t^n$ is a polynomial of degree $n - 1$ in $t$ with
leading term $nt^{n - 1}$. You can see this by thinking about what happens when
you expand all of the brackets in $(t + 1)^n$: there will be exactly one term of
degree $n$, given by picking the $t$ from every factor. There will be $n$ terms
of degree $n - 1$, because the number of ways to get a term of degree $n - 1$ is
to pick $t$ from $n - 1$ of the factors and $1$ from the other factor (so it
comes down to picking which factor you'll take the $1$ from, so you have $n$
choices). All other terms will have degree at most $n - 2$.
This argument of course comes close to proving the binomial theorem, but you
don't actually need to know anything about binomial coefficients to understand
what's going on. This slightly combinatorial flavour leads nicely into the next
part of my answer..
There is also a nice combinatorial interpretation, which is perhaps more elementary/intuitive.
Let $n \in \Bbb N$, and let $x_a^{(0)} = a^n$ and $x_a^{(m + 1)} = x_{a + 1}^{(m)} - x_a^{(m)}$. This says that the sequence $(x_1^{(m)}, x_2^{(m)}, \dotsc)$ is the sequence of $m$-th order differences of the sequence $a^n$, so it is the $m$th row (starting from $0$) of your nice ASCII diagram.
Firstly, let's interpret $x_a^{(0)} = a^n$ combinatorially. This is "the number of ways to write down an ordered list of length $n$ of numbers in $\{1, \dotsc, a\}$, with repetitions allowed".
Now let's interpret the first-order difference $x_a^{(1)} = (a + 1)^n - a^n$
combinatorially.
This is "the number of ways to write down a list of length $n$
of numbers in $\{1, \dotsc, a + 1\}$, such that $1$ appears at least once."
This is because $(a + 1)^n$ is the number of ways to write down a list of length
$n$ of numbers in $\{1, \dotsc, a + 1\}$, and $a^n$ is the number of ways to
write down a list of length $n$ of numbers in $\{2, \dotsc, a + 1\}$ (which is
obviously the same as the number of ways to write a list of numbers in
$\{1, \dotsc, a\}$).
Now similarly, the second-order difference
$x_a^{(2)} = x_{a + 1}^{(1)} - x_a^{(1)}$ is "the number of ways to
write down a list of length $n$ of numbers in $\{1, \dotsc, a + 2\}$ such that
both $1$ and $2$ appear at least once", because $x_{a + 1}^{(1)}$ is the number
of ways to write down a list of numbers in $\{1, \dotsc, a + 2\}$ with
$1$ appearing at least once, and $x_a^{(1)}$ is the number of ways to write down
a list of numbers in $\{1, 3, 4, \dotsc, a + 2\}$ with $1$ appearing at least
once
(which is obviously the same as the number of ways to write down a list of
numbers in $\{1, \dotsc, a + 1\}$ with $1$ appearing at least once).
This pattern continues, and in general we obtain that
$x_a^{(m)}$ is the number of ways to write down a list of length $n$ of numbers
in $\{1, \dotsc, a + m\}$ such that each of $1, \dotsc, m$ appears at least
once.
In particular the $n$th order difference $x_a^{(n)}$ is "the number of ways to write down a list of length $n$ of numbers in $\{1, \dotsc, a + n\}$ such that each of $1, 2, \dotsc, n$ appears at least once. Those lists are exactly the same as permutations of $\{1, \dotsc, n\}$, and there are $n!$ such permutations.
The only remotely similar answer I've been able to find is
this one, which doesn't go
into much detail, and I don't believe it interprets $x_a^{(m)}$ in general (just
for $m = n$).
The cuter interpretation of this that I gave in my comment is that
$x_a^{(m)}$ represents the number of ways you can paint $n$ houses if you have
$a + m$ different colours and you must use each of the colours $\{1, \dotsc, m\}$ at least
once. It's immediately clear then that if you must use each of the colours
$\{1, \dotsc, n\}$ at least once, the only ways you can paint the houses is by
choosing a permutation of $\{1, \dotsc, n\}$, of which there are $n!$. If you
have to use all of $\{1, \dotsc, n + 1\}$ then there is no way to paint them by
the pigeonhole principle.
It is somewhat a matter of taste if you find helps with intuition, but I think
it's a nice reason - in particular the presence of $n!$ does suggest there ought
to be a combinatorial reason for this.
