How to calculate inverse of a matrix I do not know how  to calculate the inverse of the matrix. 
\begin{vmatrix}
  1 & 1 & 1 & \cdots & 1 &\\
  \beta^{2m}_{1} &     \beta^{2m-1}_{1}  &     \beta^{2m-2}_{1}  &   \beta^{m+1}_{1} &1\\ 
   \beta^{2m}_{2} &     \beta^{2m-1}_{2}  &     \beta^{2m-2}_{2}  &   \beta^{m+1}_{2} &1 \\
\vdots & \vdots & \vdots & \ddots & \vdots &\\
   \beta^{2m}_{m-1} &     \beta^{2m-1}_{m-1}  &     \beta^{2m-2}_{m-1}  &   \beta^{m+1}_{m-1} &1 \\
 \beta^{2m}_{m} &     \beta^{2m-1}_{m}  &     \beta^{2m-2}_{m}  &   \beta^{m+1}_{m} &1
\end{vmatrix}
where $m \geq 2$ and all $\beta_i>1$ are different. 
 A: (This is not quite a solution, but it's getting close.)
We can write your matrix $A$ in partitioned form 
$$
  A = \begin{pmatrix} a^T&1\\ B&a\end{pmatrix}
$$
where $a$ is the vector with all entries equal to 1. There is a permutation matrix
$P$ such that
$$
  AP = \begin{pmatrix} 1&a^T\\ a&B\end{pmatrix}.
$$
Assume $B$ is invertible.
$$
  AP = \begin{pmatrix} 1&0\\ 0&B\end{pmatrix} \begin{pmatrix} 1&0\\ B^{-1}a&I\end{pmatrix}
          \begin{pmatrix} 1&a^T\\ 0&I-B^{-1}aa^T\end{pmatrix}.
$$
Now $\det(I-MN)=\det(I-NM)$ for any matrices $M$ and $N$ and so if $M=B^{-1}a$ and $N=a^T$,
then
$$
  \det(I-B^{-1}aa^T) = \det(1-a^TB^{-1}a) = 1 - a^TB^{-1}a.
$$
Therefore $\det(AP) =\det(B) (1-a^TB^{-1}a)$. (Here I am effectively working
out the Schur complement of $B$ in $AP$ explicitly.)
I think that in the question the matrix $B$ is a diagonal matrix times a
Vandermonde matrix, and so we can write down its inverse using Lagrnage interpolating polynomials. I do not see any
easy way to evaluate $a^TB^{-1}a$ though. 
Note that we can compute
the inverse of $AP$ using the factorization, provided we can invert $I-B^{-1}J$,
where $J=aa^T$.
For if $\beta:=a^TB^{-1}a$ and $\beta\ne1$, then it is easy to verify that
$$
  (I-B^{-1}J)(I+ \frac1{1-\beta} B^{-1}J) = I.
$$
