The following appears in BBFSK, page 118 equations (41) and (41').
$$\underset{i=1}{\overset{n}{\sum}}a_{i}\underset{k=1}{\overset{m}{\sum}}b_{k}=\underset{1\le k\le m}{\sum_{1\le i\le n}}a_{i}b_{k}$$
By complete induction we readily obtain from [the previous equation]
$$\overset{m}{\prod_{i=1}}\sum_{k=1}^{n_{i}}a_{i,k}=\underset{(i=1,\ldots,m)}{\sum_{1\leq k_{i}\leq n_{i}}}\prod_{i=1}^{m}a_{i,k_{i}};$$
where the index set for the summation on the right-hand side is the set of m-tuples $\left(k_{1},\ldots,k_{m}\right)$ with $1\leq k_{i}\leq n_{i}\left(i=1,\ldots,m\right).$
I can't make sense of the given description of the index set. It seems it should be the set of $n_{i}-tuples,$ $\left\{1,\dots,n_{i}\right\}$ with $1\le{i}\le{m}$. That is the index set is a "jagged array" of $m$ rows. Each $i^{th}$ row is $\left\{1,\dots,n_{i}\right\}$, and $n_{i}$ could be any natural number. Am I missing something, or is the description given in the book inconsistent with the expression under discussion? Is my description correct?
It helps to look at a concrete example, for example, taking $m=2$, $n_1=2$, and $n_2=3$ we get \begin{align} (a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2}+a_{2,3}) &= a_{1,1}a_{2,1} + a_{1,1}a_{2,2} + a_{1,1}a_{2,3} \\ &\qquad + a_{2,1}a_{2,1} + a_{2,1}a_{2,2} + a_{2,1}a_{2,3} \tag{1} \end{align} here we can see that all the terms on the right hand side are $2$-fold products. In your equation, an $m$-tuple $(k_1,\dots,k_m)$ corresponds to the term $\prod_{i=1}^ma_{i,k_i}$, so in $(1)$ the terms correspond to the $2$-tuples $(1,1)$, $(1,2)$, $(1,3)$, $(2,1)$, $(2,2)$, $(2,3)$ in order. This enumerates all $2$-tuples $(k_1,k_2)$ where $1\leq k_1\leq2$ and $1\leq k_2\leq3$.
In general, if you expand a product of sums $$ \prod_{i=1}^m\sum_{k=1}^{n_i}a_{i,k} \tag{2} $$ distributivity of multiplication over addition will combine every $a_{1,k_1}$ with every possible choice of $a_{2,k_2}$, then every possible choice of $a_{3,k_3}$, and so on until $a_{m,k_m}$. (This is an English interpretation of the induction you quoted in your question.) In particular, every summand in the full expansion of $(2)$ consists of $m$ factors. This is why the sum on the right hand side indexes only over $m$-tuples rather than a "jagged array" of $m$ rows. Since the distributivity makes every possible combination as described above, we have to see every possible $\prod_{i=1}^ma_{i,k_i}$ exactly once. Since each $k_i$ ranges from $1$ to $n_i$, this provides the expansion $$ \prod_{i=1}^m\sum_{k=1}^{n_i}a_{i,k} = \sum_{\substack{1\leq k_i\leq n_i \\ \forall i=1,\dots,m}}\prod_{i=1}^ma_{i,k_i} $$ as stated in your question. Therefore, the indexing set for the right hand summation is indeed the set of all $m$-tuples $(k_1,\dots,k_m)$ where $1\leq k_i\leq n_i$ for every $i=1,\dots,m$.