I'm trying to understand the steps in a certain proof, where I think the following claim is used: $S( |i \rangle \langle i| \otimes \rho) = S(\rho)$, where $S()$ is the entropy, $|i \rangle$ is a state from some orthonormal basis, and $\rho$ is a density operator. Here's my attempt to see why this holds:
$$ \begin{align} S( |i \rangle \langle i| \otimes \rho) &= -tr\left[\left(|i \rangle \langle i| \otimes \rho\right) \times \log \left(|i \rangle \langle i| \otimes \rho\right)\right]\\ &= -tr[(|i \rangle \langle i| \times \log |i \rangle \langle i|) \otimes \rho \log \rho]\\ &= -tr[(|i \rangle \langle i| \times \log |i \rangle \langle i|)] \times tr[\rho \log \rho]\\ &= 0 \times S(\rho) \end{align}$$
so I am clearly making a mistake somewhere, that $0$ should not be there. The first equality comes from the definition of $S()$, the second because $(A \otimes B)(C \otimes D) = AC \otimes BD$, the third because $tr(A \otimes B) = tr(A) \times tr(B)$. Are any of these incorrect?