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.