Tensor product properties used to obtain Kraus operator decomposition of a channel

I work on a Quantum Information Science II: Quantum states, noise and error correction MOOC by Prof. Aram Harrow, and I do not understand which property of tensor products is used in one of the transitions in the videos.

Let’s consider an isometry $V: A to B otimes E$ ($E$ is a subspace to be thrown away at the end).

Let’s fix and orthonormal basis ${ |erangle }$ in $E$ and partially expand the isometry $V$ as $V = sum_e V_e otimes |erangle$, where each $V_e$ is a linear operator from $A$ to $B$.

The Stinespring form of a quantum operation is a partial trace applied after an isometry: $N(rho) = mathrm{tr}_E [V rho V^dagger]$.

Now, if we expand that with our representation of $V$, we get
$$
N(rho) = mathrm{tr}_E left[
sum_{e_1} sum_{e_2}
left( V_{e_1} otimes |e_1rangle right)
rho
left( V_{e_2}^dagger otimes langle e_2| right)
right].
$$

My question is how to get from here to the next step
$$
N(rho) = mathrm{tr}_E left[
sum_{e_1} sum_{e_2}
(V_{e_1} rho V_{e_2}^dagger) otimes |e_1 rangle langle e_2|
right]?
$$

(BTW, eventually, we end up with the Kraus operator decomposition of a channel: $N(rho) = sum_e V_e rho V_e^dagger$.)