Inverse of a specific matrix Let $A$ be an $(n+1)\times(n+1)$ matrix defined by $a_{ij} = (i-1)^{j-1}$, with the convention that $0^0 = 1$.
$$A = 
\left[\begin{matrix}
0^0 & 0^1 & 0^2 & \ldots & 0^n\\
1^0 & 1^1 & 1^2 & \ldots & 1^n\\
2^0 & 2^1 & 2^2 & \ldots & 2^n\\
\vdots & \vdots & \vdots &\ddots & \vdots\\
n^0 & n^1 & n^2 & \ldots & n^n\\
\end{matrix}\right]
$$ 
What is a closed form of $A^{-1}$ ?    
I don't know how to approach this problem.
Any help will be appreciated.
 A: Your matrix is a type of Vandermonde matrix.
If we denote $x_i=i$ and recall that $x_0=0$, we can write
$$
A^T
=
\begin{pmatrix}
1&1&\dots&1\\
0&x_1^1&\dots&x_{n}^1\\
\vdots&\vdots&&\vdots\\
0&x_1^n&\dots&x_{n}^n
\end{pmatrix}
=
\begin{pmatrix}
1&u\\
0&V
\end{pmatrix},
$$
where $u=(1,\dots,1)$ is the row vector of all ones and
$$
V
=
\begin{pmatrix}
x_1^1&\dots&x_{n}^1\\
\vdots&&\vdots\\
x_1^n&\dots&x_{n}^n
\end{pmatrix}.
$$
As long as all the $x_i$s are different, this $V$ has an explicit inverse.
See this ProofWiki entry for details.
The key element is $B:=V^{-1}$ given behind that link.
Using it and
$$
\begin{pmatrix}
1&u\\
0&V
\end{pmatrix}^{-1}
=
\begin{pmatrix}
1&-uV^{-1}\\
0&V^{-1}
\end{pmatrix},
$$
we get
$$
A^{-1}
=
\begin{pmatrix}
1&0\\
-B^{T}u^T&B^{T}
\end{pmatrix}.
$$
To make this more explicit, you need to take the actual formula for $B=V^{-1}$.
Its elements (see the link above) are
$$
b_{i j} = \begin{cases}
\left({-1}\right)^{j-1} \left({\dfrac{\displaystyle \sum_{\substack{1 \mathop \le m_1 \mathop < \ldots \mathop < m_{n-j} \mathop \le n \\ m_1, \ldots, m_{n-j} \mathop \ne i} } x_{m_1} \cdots x_{m_{n-j}} } {x_i \displaystyle \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne i} } \left({x_m - x_i}\right)}}\right) & \text{when } 1 \le  j < n \\
\qquad \qquad \qquad \dfrac 1 {x_i \displaystyle \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne i} } \left({x_i - x_m}\right)} & \text{when } j = n.
\end{cases}
$$
As $x_i=i$, you can simplify this much further.
We can first observe that
$$
\prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne i} } \left({x_i - x_m}\right)
=
\prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne i} } (i - m)
=
(i-1)!
(n-i)!
(-1)^{n-i},
$$
as this is the product of consecutive integers skipping over zero.
To get further, we need to understand the sum
$$
S_j
=
\sum_{\substack{1 \mathop \le m_1 \mathop < \ldots \mathop < m_{n-j} \mathop \le n \\ m_1, \ldots, m_{n-j} \mathop \ne i} } m_1 \cdots m_{n-j}
.
$$
Some extreme cases are simple enough to see.
Clearly
$$
S_1
=
1\cdot2\cdots(i-1)\cdot(i+1)\cdots n
=
n!i^{-1}
$$
and
$$
S_{n-1}
=
1+2+\cdots(i-1)+(i+1)+\cdots+n
=
\frac12n(n+1)-i.
$$
I can't see how to get a general closed form expression.
It is probably possible, but will not be very neat.
One thing that is going to be easier to handle is the vector $w=B^Tu^T$.
Its components are given by $w_i=\sum_{j=1}^nb_{ji}$.
But clearly a neater closed form expression for $b_{ij}$ will give $w_i$ as well, so I guess finding only $w_i$ would be unsatisfactory.
