How to prove that the inverse of following matrix exist and has integer entries?

Question

How to prove that the following matrix has an inverse with integer entries? Also find out the inverse.

\begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{3} & \cdots &\ \frac{1}{n-1} & \frac{1}{n} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \cdots &\ \frac{1}{n} & \frac{1}{n+1} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \cdots &\ \frac{1}{n+1} & \frac{1}{n+2} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \cdots &\ \frac{1}{n+1} & \frac{1}{n+2} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\ \frac{1}{n} & \frac{1}{n+1} & \frac{1}{n+2} & \cdots &\ \frac{1}{2n-2} & \frac{1}{2n-1} \\ \end{bmatrix}

Answer 1

I am assuming that the question is about the matrix $A$ whose $(i,j)$ entry is $A_{ij}=\frac{1}{i+j-1}$. (As remarked in the comments, there is a typo as written in the question.)

I claim that $A^{-1}$ has the following explicit formula: $$ \left(A^{-1}\right)_{ij}=(-1)^{i+j}\binom{n+i-1}{n-j}\binom{n+j-1}{n-i}\binom{i+j-2}{i-1}^2(i+j-1). $$

To verify that this is indeed the inverse, I would suggest using induction to show that its product with $A$ is the identity matrix.