Prove the inverse of the Hilbert matrix has integer entries 
$1 \frac{1}{2} ... \frac{1}{n}$
$\frac{1}{2} \frac{1}{3} ... \frac{1}{n+1}$
$.$
$.$
$.$
$\frac{1}{n} \frac{1}{n+1} ... \frac{1}{2n-1}$
Does the inverse of this matrix have integer entries? Prove your
  statement.

I thought of  multiplying every line $i$ by $i!$, but it leads to a very complicated solution (if it leads to a solution at all).
Source: Linear Algebra, Kenneth Hoffman and Ray Kunze. Section 1.6, exercise 12.

The answer to your first question is yes.
Do not dismiss the example referenced in the problem statement (the case where $n = 3$). Examples like the one I'll reproduce below help to build insight:

Given $\quad A=\begin{pmatrix}1 & \frac{1}{2} & \frac{1}{3} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5}\end{pmatrix},\qquad$
  $A^{-1}=\begin{pmatrix} 9 & -36 & 30 \\ -36 & 192 & -180 \\ 30 & -180 & 180 \end{pmatrix}$.

I'd suggest you explore the matrix and its inverse for $n = 4$.
What we want to show is that
$B=\begin{pmatrix} 1 & \frac{1}{2} & \ldots & \frac{1}{n} \\ \frac{1}{2} & \frac{1}{3} & \ldots & \frac{1}{n+1} \\ \ldots & \ldots & & \ldots \\ \frac{1}{n} & \frac{1}{n+1} & \ldots & \frac{1}{2n-1}\end{pmatrix}$
is invertible and $B^{-1}$ has integer entries. 
Hints to get you started: The matrix $B$ is known as a Hilbert matrix and the entries of its inverse can be represented as the product of binomial coefficients. 
