Some questions about Hilbert matrix This Exercise $12$ page $27$ from Hoffman and Kunze's book Linear Algebra.

The result o Example $16$ suggests that perhaps the matrix
$$A = \begin{bmatrix} 1 & \frac{1}{2} & \ldots & \frac{1}{n}\\
 \frac{1}{2} & \frac{1}{3} & \ldots & \frac{1}{n+1}\\ \vdots & \vdots &
 & \vdots \\ \frac{1}{n} &\frac{1}{n+1} &\ldots &\frac{1}{2n-1}
\end{bmatrix}$$
is invertible and $A^{-1}$ has integer entries. Can you prove that?

I will copy their words before that example: 

Some people find it less awkward to carry along two sequences of
  matrices, one describing the reduction of $A$ to the identity and the
  other recording the effect of the same sequence of operations starting
  from the identity.

In this post, Adrián Barquero says: 

If your linear algebra textbook is Kenneth Hoffmann and Ray Kunze's
  Linear Algebra book then I was in the same situation you're right now
  a few years ago, but I was adviced at the moment that this particular
  exercise wasn't as easy as one might expect at first.
The matrix you have is called the Hilbert matrix and the question you
  have was already asked a couple of times in math overflow here and
  here. They have excellent answers so I will just point you to them.

My question: Is it possible answer the question made by the authors without using high techniques?  I am not allowed to use determinants here.
PS: Those answers in MO don't answer the question above. I apologise if this question is unappropriated or if I was unable to understand their answers.
 A: Yes, I believe a student at the level of this book is able to answer the question. But this is not easy. I think it would be an issue at the Mathematical Olympiad.
I suggest you do this proof by induction on the size $n$ of the $n \times n$ Hilbert matrix $A_n.$ Start the induction step with $n = 2 $ . The statement is essentially the following:
$$
 A_n = \left(\frac{1}{i + j-1}\right)_{ij} \mbox{ is a Hilbert matrix
}\Rightarrow A_n^{-1} \in M_{n \times n}(\mathbb{Z}).
 $$
Step 1. Verify that by inspection for $ n = 2 $.
Step 2. Verify that if $A_n$ is a Hilbert matrix, then
$$
A_{n+1}=
\begin{pmatrix} 
A_n & v^{T} \\
v   & \frac{1}{2(n+1)-1}
\end{pmatrix}
\quad
$$
is a Hilbert matrix with $v=\left(\frac{1}{n+1},\dots,\frac{1}{2(n+1)-2}\right)$. Now apply matrix inversion in block form, special case 1.
And in this special case in which the blocks are matrices that commute, there is an exercise in the Hoffman book (do not remember which page).
Edit 1.
And in this particular case, this formula can be obtained using the definition of the inverse of a matrix. Just partition the inverse matrix into blocks of the same size and make the product and solve a matrix system. I believe that a student at the Mathematical Olympiad may have this idea, and an average student can understand it.
Edit 2. Done what was said in Edit 1, according to the notation of the link, just to verify that for $k =\frac{1}{2(n+1)-1} - vA_n^{-1}v^T,$ we have that $ \frac{1}{k}A_n^{-1}v^T \in\mathbb{Z} $. But this can be checked again by induction.
