Show that the eigenvalues of $A = QTQ^\ast$ are the diagonal entries of $T$?

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$Q$ is a unitary $n\times n$ matrix of complex numbers. $T$ is an upper triangular $n$ x $n$ matrix of complex numbers. $Q^\ast$ is the complex-conjugate transpose of $Q$.

I've tested some sample matrices in MATLAB, but it didn't really give me an understanding of why this holds true. Looking at $A$, there didn't appear to be any noticeable property other than the diagonals all being real numbers, which I don't know is always true.

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As pointed out in the comments, by properties of the determinant and the fact that here $Q^*=Q^{-1}$ we have \begin{equation}\det{A}=\det{(QTQ^*)}=\det {Q}\det{T}\det {Q^*}=\det{T}.\end{equation} Since $T $ is (upper) triangular, its determinant equals the product of its diagonal entries.