Let's say $A$ is a diagonal-constant matrix with $1$ on the main diagonal and $2$ on the first superdiagonal and other element are zero .
$$A= \begin{bmatrix} 1& 2& 0& 0& 0&0& ...&0 \\ 0& 1& 2& 0& 0& 0& ...& 0\\0&0&1&2&...\\...&&&&&&&2\\0&0&0&...&&&&1 \end{bmatrix}$$
It is obvious determinant of $A$ is $1$. By manual calculation I can see that
$$A^{-1} = \begin{bmatrix} 1& -2& 4& &&&&-2^{m-1} \\0& 1& -2& 4& &&&2^{m-2}\\ &&\\&&&\\\\0&0&0&&&1&-2&4\\0&0&0&&&&1&-2\\0&0&0&...&&&&1 \end{bmatrix}$$
Is there a way to prove what the inverse of $A$ would be?
If $B$ is a matrix for which the series $\sum_{k=0}^\infty B^k$ converges, then we necessarily have $$ \sum_{k=0}^\infty B^k = (I - B)^{-1} $$ as you can verify by calculation. Your matrix is an example of this, where we take $$ B = -\pmatrix{0&2\\&0&2\\&&\ddots & \ddots\\&&&0&2\\&&&&0} $$ To see that the sum converges, it suffices to note that $B^k = 0$ whenever $k$ is sufficiently large.
Note, however, there is no general formula which makes it easier to take the inverse of a matrix with a constant diagonal. The formula I presented will work, however, whenever the matrix is triangular (upper triangular or lower triangular) with a constant diagonal, since all strictly triangular matrices are nilpotent.