An identity for the factorial function A friend of mine was doodling with numbers arranged somewhat reminiscent of Pascal's Triangle, where the first row was $ 1^{n-1} \ \  2^{n-1} \ \cdots \ n^{n-1} $ and subsequent rows were computed by taking the difference of adjacent terms. He conjectured that the number we get at the end is $ n! $ but I've not been able to prove or disprove this. The first few computations are given below:
$$ 
    \begin{pmatrix}
    1 \\
    \end{pmatrix}
$$
$$
    \begin{pmatrix}
    1   &     & 2   \\
        & 1   &     \\
    \end{pmatrix}
$$
$$
        \begin{pmatrix}
        1   &     & 4   &     & 9   \\
            & 3   &     & 5   &     \\
            &     & 2   &     &     \\
        \end{pmatrix}
$$
$$
        \begin{pmatrix}
        1   &     & 8   &     & 27  &     & 64  \\
            & 7   &     & 19  &     & 37  &     \\
            &     & 12  &     & 18  &     &     \\
            &     &     & 6   &     &     &     \\
        \end{pmatrix}
$$
$$
\newcommand\pad[1]{\rlap{#1}\phantom{625}}
        \begin{pmatrix}
        1   &     & 16  &     & 81  &     & 256 &     & 625 \\
            & 15  &     & 65  &     & 175 &     & 369 &     \\
            &     & 50  &     & 110 &     & 194 &     &     \\
            &     &     & 60  &     & 84  &     &     &     \\
            &     &     &     & 24  &     &     &     &     \\
        \end{pmatrix}
$$
I attempted to write down the general term and tried to reduce that to the required form. The general term worked out as
$$
       \sum_{i=0}^n (-1)^{n-i} \binom{n}{i} (i+1)^{n}.
$$
I tried applying various identities of the binomial coefficients but I'm barely making any progress. Any help would be appreciated.
Small note: If I instead start with the first row as $ 0^{n} \ \  1^{n} \ \cdots \ n^{n} $ then I still get $n!$ at the end of the computation, and the general formula in this case works out as
$$
       \sum_{i=0}^n (-1)^{n-i} \binom{n}{i} i^{n}.
$$
In fact, we can start with any $n$ consecutive natural numbers, each raised to the $(n-1)$th power, and we still get $n!$ at the end of the computation.

Let's say we want to prove that iterating $n^3$ three times gives us $3!$. (Assume we already know that your thing works for $n^2$, $n$, and $1$.)
Well, if you iterate $n^3$ once, you get $(n+1)^3-n^3=3n^2+3n+1$. We also know that iterating $3n^2$ two times gives us $3\times2!=3!$, and we know that iterating $3n$ and $1$ two times both give us $0$ (since they become constant after one or zero iterations respectively).
Thus, iterating $3n^2+3n+1$ two times gives us $3!$, and therefore iterating $n^3$ three times gives us $3!$.
Can you generalize this argument?
