Summation Notation Question in McMillan's Theorem Proof Let me preface by saying that this question does not pertain as much to coding theory, as it does to mathematical notation. Every symbol in this question is a natural number.
Anyhow, I am currently reading through a proof of McMillan's Theorem. In the proof, there is the statement that
\begin{equation}\tag{1}
\left( \sum_{j=1}^m \dfrac{\alpha_j}{r^j} \right)^u = \left( \dfrac{\alpha_1}{r} + \cdots + \dfrac{\alpha_m}{r^m}  \right)^u, 
\end{equation}
and so far, so good. However, the next line states that $(1)$ out can be multiplied out to yield
\begin{equation}\tag{2}
\sum_{\substack{1 \leq i_j \leq m \\ i_1, i_2, \cdots, i_u}} \dfrac{\alpha_{i_1}}{r^{i_1}} \cdots \dfrac{\alpha_{i_u}}{r^{i_u}} = \sum_{\substack{1 \leq i_j \leq m \\ i_1, i_2, \cdots, i_u}} \dfrac{\alpha_{i_1} \alpha_{i_2} \cdots \alpha_{i_u}}{r^{i_1 + \cdots + i_u}},
\end{equation}
which confuses me. Can someone explain to me what the subscripts on the sigmas in equation $(2)$ are meant to indicate?
Also, I apologize if this question has already been answered somewhere else. I tried to look through the archives and didn't come across it.

Would this help?
\begin{align}
    \left( \sum_{j=1}^2 \dfrac{\alpha_j}{r^j} \right)^3 
    &= \left( \dfrac{\alpha_1}{r}+\dfrac{\alpha_2}{r^2} \right)^3 \\
    &= \left( \dfrac{\alpha_1}{r^1}+\dfrac{\alpha_2}{r^2} \right)
       \left( \dfrac{\alpha_1}{r^1}+\dfrac{\alpha_2}{r^2} \right)
      \left( \dfrac{\alpha_1}{r^1}+\dfrac{\alpha_2}{r^2} \right) \\
    &= \dfrac{\alpha_1}{r^1}\dfrac{\alpha_1}{r^1}\dfrac{\alpha_1}{r^1}+\\
       &\phantom{=} \left(
          \dfrac{\alpha_1}{r^1}\dfrac{\alpha_1}{r^1}\dfrac{\alpha_2}{r^2}+
          \dfrac{\alpha_1}{r^1}\dfrac{\alpha_2}{r^2}\dfrac{\alpha_1}{r^1}+
          \dfrac{\alpha_2}{r^2}\dfrac{\alpha_1}{r^1}\dfrac{\alpha_1}{r^1}
      \right) +\\
      &\phantom{=} \left(
          \dfrac{\alpha_1}{r^1}\dfrac{\alpha_2}{r^2}\dfrac{\alpha_2}{r^2}+
          \dfrac{\alpha_2}{r^2}\dfrac{\alpha_1}{r^1}\dfrac{\alpha_2}{r^2}+
          \dfrac{\alpha_2}{r^2}\dfrac{\alpha_2}{r^2}\dfrac{\alpha_1}{r^1}
          \right) + \\
       &\phantom{=}
        \dfrac{\alpha_2}{r^2}\dfrac{\alpha_2}{r^2}\dfrac{\alpha_2}{r^2}\\
    &= \dfrac{\alpha_1\alpha_1\alpha_1}{r^{1+1+1}} + \\
    &\phantom{=} \left(
       \dfrac{\alpha_1\alpha_1\alpha_2}{r^{1+1+2}}+
       \dfrac{\alpha_1\alpha_2\alpha_1}{r^{1+2+1}}+
       \dfrac{\alpha_2\alpha_1\alpha_1}{r^{2+1+1}}
   \right) +\\
   &\phantom{=} \left(
       \dfrac{\alpha_1\alpha_2\alpha_2}{r^{1+2+2}}+
       \dfrac{\alpha_2\alpha_1\alpha_2}{r^{r^2+1+2}}+
       \dfrac{\alpha_2\alpha_2\alpha_1}{r^{r^2+2+1}}
       \right) + \\
    &\phantom{=} \dfrac{\alpha_2\alpha_2\alpha_2}{r^{2+2+2}} \\
\end{align}
