Inverse of a special Vandermonde Matrix with elements $i^{j-1}$

Question

I have a $(n+1) \times (n+1)$ Vandermonde matrix $$ V_n= \begin{pmatrix} 1 & 1 & \dots & 1\\ 1 & 2 & \dots & 2^n\\ \vdots & \vdots & \ddots & \vdots\\ 1 & n+1 & \dots & (n+1)^n\\ \end{pmatrix} $$ and by observation (and lots of oeis.org searching) I suspect that $$ V_n^{-1}=(b_{i,j}) $$ with $$ b_{i,j}=\frac{(-1)^{i+j}}{n!}\binom{n}{j-1}\sum_{m=0}^{n+1-i} (-j)^{m}{ n+2\brack i+1+m} $$ where ${ n\brack m}$ are unsigned Stirling numbers of the first kind. Now I wonder how to prove this?

My attempt

I am aware there is a Matrix inverse formula for Vandermonde matrices, for example mentioned in this post, but I failed to match it to the formula above. Another approach where I got farther was to use the definition of matrix multiplication for $V_{n}^{-1}V_n=(c_{i,j})$ and see whether it simplifies to the identity matrix. Here I've got after some algebra mess (it is lengthy to include here, but I can add if needed) to $$ c_{i,j}=(-1)^{n+i-j}\sum_{k=n}^{n-i+j}(-1)^{k}{ n+2\brack k+i-j+2} {k+1 \brace n+1}. $$ (here ${n \brace m}$ are Stirling numbers of the second kind for a change...) From this we can see that for $j<i$ the sum is empty and we get a zero under the diagonal. For $i=j$ we get $c_{i,i}=1$ on the diagonal. So to finish the proof it needs to be shown there are zeroes above the diagonal, i.e. $c_{i,j}=0$ for $i<j$, and that is where I could not get any further.

Can we prove the last step somehow? Or is there another way to prove the inverse identity?

Answer 1

Use formula

$$V_n=S_nD_nP_n^T \tag{1}$$

which implies:

$$V_n^{-1}=(P_n^{-1})^T D_n^{-1}\underbrace{S_n^{-1}}_{s_n}\tag{2}$$

where $s_n$, resp. $S_n$, are the lower triangular matrices with Stirling numbers of the first, resp. second kind, $D_n=diag(1!, 2!, \cdots n!)$, and $P_n$ the lower triangular Pascal matrix.

Formula (1) is detailed in lemma 2 of this very interesting article entitled "On a connection between the Pascal, Stirling and Vandermonde matrices" by Yang and You (not Me and You). This article is mentionned in particular here where you will find other interesting informations.

It must be added that the inverse of Pascal matrix is easy to describe : it is obtained from $P_n$ by multiplying its entries $a_{ij}=\binom{i-1}{j-1}$ by $(-1)^{i+j}$.

From (2), can you retrieve your formula ? (for example the $(-1)^{i+j}$ comes from the above mentionned inversion of $P_n$).

Remark: Your attempt looks like verifying that $S_ns_n=I$, a property we have used above.