Partial trace of matrix product state

Question

I have come accross a formula that puzzles me a bit in the proof of lemma 23 (page 32) of [this paper][1].

The authors start from a (translationally-invariant) matrix product state:
$$\lvert\psi\rangle := \sum_{i_1, \ldots, i_n}\textrm{tr}(A_{i_1} \ldots A_{i_n})\lvert i_1 \ldots i_n\rangle$$

where the $A_i$ are matrices of dimension $D$ satisfying $\sum_{i}A_i^{\dagger}A_i = \sum_{i}A_iA_i^{\dagger} = \mathbf{1}_{D \times D}$ (that is, they constitute Kraus operators for a unital channel). Then, they claim that the reduced state of the first $l$ qudits looks like:

$$\rho_l = \sum_{i_1, \ldots, i_l}\textrm{tr}\left(A_{i_1} \ldots A_{i_l}\frac{\mathbf{1}_{D \times D}}{D}A_{j_1}^{\dagger} \ldots A_{j_l}^{\dagger}\right)\lvert i_1 \rangle\langle j_1\rvert \otimes \ldots \otimes \lvert i_l \rangle\langle j_l \rvert$$

Yet, the first expression that looks obvious to me would rather be:
$$\rho_l = \sum_{i_1, \ldots, i_l}\left(\sum_{i_{n + 1}, \ldots, i_n}\textrm{tr}\left(A_{i_1}\ldots A_{i_{l}}A_{i_{l + 1}}\ldots A_{i_n}\right)\textrm{tr}\left(A_{i_n}^{\dagger}\ldots A_{i_{l + 1}}^{\dagger}A_{i_l}^{\dagger}\ldots A_{i_1}^{\dagger}\right)\right)\lvert i_1 \rangle\langle j_1\rvert \otimes \ldots \otimes \lvert i_l \rangle\langle j_l \rvert$$

It looks as if by some magic the authors converted the sum of product of traces above to the trace of a product! Any idea where that would come from? At first, I thought it had to do with the $A$ being randomized (see paragraph on top of page 32), but by giving it a closer look, it seems the statements of lemma 23 do not include an expectation value with respect to the Haar measure, as opposed e.g to lemma 24, so I am not sure what to think.

  [1]: https://arxiv.org/abs/1206.2947