I think the claim in this form is wrong. The eigenvalues are not integral. For example, with $n=4$ the matrix is
$$
A=\begin{pmatrix}
0 & 4 & 0 & 0\cr
1 & 0 & 3 & 0\cr
0 & 2 & 0 & 2 \cr
0 & 0 & 3 & 0
\end{pmatrix}.
$$

The characteristic polynomial of this matrix is 
$\chi (t)=t^4-16 t^2 +24$, which has no integral roots.
Am I overlooking something ?

Edit: just visited the site http://math.nist.gov/MatrixMarket/deli/Clement/
Here I saw that the upper diagonal must be $3,2,1$, not $4,3,2$. Then everything is OK.
So $y_k=n-k$ rather than $y_k=n-k+1$, what you wrote. We have a recursion for $\chi(t)=\det(t\cdot id-A)$, see http://mathoverflow.net/questions/131527/eigenvalues-of-symmetric-tridiagonal-matrices.

Edit: A proof for the correct claim can be found in the paper of Taussky and Todd, "Another look at a matrix of Mark Kac", Linear Algebra Appl. 150 (1991), 341-360.
More details are also at
http://math.stackexchange.com/questions/405670/regarding-a-paper-by-paul-a-clement-on-tridiagonal-matrices, which was remarked by Gerry (thank you ). Also, Darij's remark is correct, that the proof does not follow easily, so I have edited this here.