This is exercise 10.2.12 in Dummit and Foote.
Problem: Let $I$ be a left ideal of $R$ and $n$ a positive integer. Show that $$ R^n/IR^n \cong R/IR \times \dots \times R/IR $$ where RHS is product $n$-times.
The preceding example has you prove that if $A_1,\dots A_n$ are modules with $B_1,\dots,B_n$ corresponding submodules, that $$ (A_1 \times \dots \times A_n)/(B_1 \times \dots \times B_n) \cong (A_1/B_1) \times \dots \times (A_n/B_n). $$ This is fine to prove, so all you really need to do is show that $IR^n = (IR)^n$ in order to prove the original problem, because then it follows from the previous exercise. The issue I'm having is actually showing the equality. One direction is more straightforward than the other, but maybe I'm over complicating things?
Attempt: Let $x \in IR^n$ then $x = \sum_1^ma_i(r_{i1},r_{i2},\dots,r_{in})$ where $a_i \in I$ and $(r_{i1},\dots,r_{in}) \in R^n$. Then we can collect the summands to get $$ x = \left(\sum_1^ma_ir_{i1},\sum_1^ma_ir_{i2},\dots,\sum_1^ma_ir_{in} \right) $$ which is exactly the form of elements in $(IR)^n$ since $IR = \{a_1r_1+\dots+a_nr_n \mid n \geq 1, a_i \in I, r_i \in R\}$.
The issue I'm having is doing the reverse inclusion. Because if $x \in (IR)^n$, I believe all I can say is that $$ x = \left( \sum_1^m a^{(1)}_ir^{(1)}_i, \dots, \sum_1^m a^{(n)}_ir^{(m)}_i \right). $$ Even here I've tacitly assumed that the sums are all of the same length, which I believe is alright because we could take the maximum and then fill in the rest with $0$ coefficients wlog. I saw another answer online that said an element of $(IR)^n$ would have form $(a_1r_1,\dots,a_nr_n)$ for $a_i \in I$ but I don't see why this would be the case? Is there a way to do this without all of the double indicies also? If there's a much simpler way to do this please let me know. Thanks in advance.
Following the hint given by Mark, we can proceed as follows.
If $x \in (IR)^n$ then $x = (x_1,\dots,x_n)$ with $x_i \in IR$ for all $1 \leq i \leq n$. Notice we can decompose $x$ as $$ x = (x_1,\dots,0) + (0,x_2,\dots,0) + \dots (0,\dots,x_n), $$ and so if we show that an arbitrary one of these components is in $IR^n$ then we will have that $x \in IR^n$ since as a module $IR^n$ is closed under finite sums. Fix some $1 \leq i \leq n$ and consider the tuple $(0,\dots,x_i,\dots,0)$. Then $x_i = \sum_1^ma_ir_i$ where $a_i \in I$ and $r_i \in R$. It follows that \begin{align*} (0\dots,x_i,\dots,0) &= \left(0,\dots,\sum_1^ma_ir_i,\dots,0\right) \\ &= \sum_1^m(0,\dots,a_ir_i,\dots,0) \\ & = \sum_1^ma_i(0,\dots,r_i,\dots,0). \end{align*} It is obvious that $a_i(0,\dots,r_i,\dots,0) \in IR^n$ and so $(0,\dots,x_i,\dots,0) \in IR^n$ from which it follows that $x \in IR^n$ since the module is closed under finite sums. Therefore $IR^n = (IR)^n$.