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 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.