We work inside a saturated model $M$ of a complete theory $T$ in some language $L$. Let $A \subseteq M$ and $(b_i , i < \lambda)$ a "very long" sequence such that $\operatorname{tp}(b_0) = \operatorname{tp}(b_i)$ for all $i$. By Erdös-Rado, we may assume $(b_i)$ is $A$-indiscernible.
I am having trouble with this remark made in class, as I'm not sure how does Erdös-Rado applies here. My first thought was that this is just an application of the "Standard Lemma" for indiscernible sequences, but it doesn't quite allow to just rename the sequence as the new sequence obtained might be completely unrelated.
I also tried to use the following lemma from Tent & Ziegler:
Lemma (7.2.11): For all $A$ there is some $\lambda$ such that for any linear order $I$ of size $\lambda$ and any family $(a_i, i \in I)$ there is an $A$-indiscernible sequence $(b_j , j < \omega)$ such that for all $j_1 ,\dots, j_n < \omega$ there is a sequence $i_1,\dots,i_n \in I$ such that $ \operatorname{tp}(j_1,\dots,j_n/A) = \operatorname{tp}(i_1,\dots,i_n/A)$.
The proof of this lemma does use Erdös-Rado (although non-trivially).I figured that since $\operatorname{tp}(b_0) = \operatorname{tp}(b_i)$, we might reproduce said proof without changing the sequence altogether (aside from reindexing and concatenating some of the $b_i$'s ). Maybe it's just a more direct application of the theorem, but I'm not very used to these combinatorial principles.