Is there a shorter way to prove this? Let $n$ a natural number and $A=(a_{ij})$, where $a_{ij}=\left(\begin{array}{c}i+j \\i\end{array} \right)$, for $0≤i,j<n$. Prove that A has an inverse matrix and that all the entries of $A^{−1}$ are integers.
I tried to prove this like this: (but I'm not sure if it's correct or formal enough)


Also if you could give me a hint to prove the second aasertion it would be great. 
Thank you.
 A: There are (at least) two approaches to showing that the inverse has integer entries.  The first is Cramer's rule, which gives the entries of the inverse of $A$ as quotients of the form "determinants of certain submatrices of $A$ divided by the determinant of $A$".  Since the entries of your $A$ are integers, and since your row-reduction shows that the determinant of $A$ is $1$, it follows that these quotients are integers.
Alternatively, do you know the algorithm for inverting a square matrix by (1) writing an identity matrix of the same size next to it (so that the combination is an $n\times 2n$ matrix), (2) applying elementary row operations to convert the original matrix into the identity matrix, and (3) reading off the inverse in the region where you originally put the identity matrix?  If so, then you've done most of the work for showing that the inverse of your matrix has integer entries.  You've row-reduced your matrix to upper triangular form with $1$'s (not just non-zero entries, but $1$'s) on the diagonal, without ever having to do any dividing.  You could complete the row-reduction to the identity without any division.  The whole reduction process, if applied to an identity matrix, would never produce anything but integers.
A: Let $B=\left[b_{ij}=\binom{i}{j}\right]$ and $C=BB^t$. We have $c_{ij}=\sum_{k=0}^i\binom{i}{k}\binom{j}{k}=\binom{i+j}{i}=a_{ij}$, thus $C=A$. So it suffices to show that the inverse of $B$, has all integer entries. Because $B$ is lower-triangular $B^{-1}$ is also lower-triangular and can be found column by column. The first column of $B^{-1}$ is a vector of alternating  $+1$ and $-1$s. I think it should be easy from this point to show the entries in the next columns are also integers one by one. To be precise you need to inspect the linear equations corresponding to the entry of $B^{-1}$ being inspected.
