A $\{0,1\}$-matrix with positive spectrum must have all eigenvalues equal to $1$

Here's a cute problem that was frequently given by the late Herbert Wilf during his talks.

Problem: Let $A$ be an $n \times n$ matrix with entries from $\{0,1\}$ having all positive eigenvalues. Prove that all of the eigenvalues of $A$ are $1$.

Proof:

Use the AM-GM inequality to relate the trace and determinant.

Is there any other proof?

If one wants to use the AM-GM inequality, you could proceed as follows: Since $A$ has all $1$'s or $0$'s on the diagonal, it follows that $tr(A)\leq n$. Now calculating the determinant by expanding along any row/column, one can easily see that the determinant is an integer, since it is a sum of products of matrix entries (up to sign). Since all eigenvalues are positive, this integer must be positive. AM-GM inequality implies $$det(A)^{\frac{1}{n}}=\left(\prod_{i}\lambda_{i}\right)^{\frac{1}{n}}\leq \frac{1}{n}\sum_{i=1}^{n}\lambda_{i}\leq 1.$$ Since $det(A)\neq 0$, and $m^{\frac{1}{n}}>1$ for $m>1$, the above inequality forces $det(A)=1$. We therefore have equality which happens precisely when $\lambda_{i}=\lambda_{j}$ for all $i,j$. Combining this with the above equality gives the result.