Generators of a direct product of symmetric groups This question is a generalization of Minimum size of the generating set of a direct product of symmetric groups to a product of arbitrary arity, and the only answer to that question is incomplete (it doesn't talk about $S_m \times S_n$ with both $m$ and $n$ even).
So, the question is: how can one build a minimal set of generators for the direct product $S_{n_1} \times S_{n_2} \times \ldots \times S_{n_k}$?
 A: Q. how can one build a minimal set of generators for the direct product
$S_{n_1}\times \cdots \times S_{n_k}$?
I will interpret minimal to mean minimal size. 
First, there is a unique minimal size set of generators for $S_2$, and it is
$\{(1\;2)\}$. It has size $1$.
For $n>2$, a minimal size set of generators for $S_n$ has size $2$,
and $\{(1\;2), (1\;2\;\cdots\;n)\}$ is one such. Another $2$-element generating set is $\{(1\;2), (2\;3\;\cdots\;n)\}$.
There is nothing more that needs to be said about the case
where the number of factors, $k$, is $1$.
I claim that, when $k>1$ and all factors are nontrivial,
a minimal size set of generators for
$S_{n_1}\times \cdots \times S_{n_k}$ has size $k$.
$k$ is a lower bound.
$(S_{n_1}\times \cdots \times S_{n_k})/
(A_{n_1}\times \cdots \times A_{n_k})\cong \mathbb Z_2^k$,
which is a vector space over the $2$-element field.
This quotient requires $k$ generators,
because it is $k$-dimensional, so the original group
$S_{n_1}\times \cdots \times S_{n_k}$
requires at least $k$ generators.
An explicit generating set with no more than $k$ elements.
I will follow the idea of
Minimum size of the generating set of a direct product of symmetric groups, where this problem is solved for the case $k=2$.
I assume that $S_{n_i}$ acts on the set
$X^{(i)} = \{1^{(i)},\ldots,n_i^{(i)}\}$,
and that the sets $X^{(i)}$, $1\leq i\leq k$,  are pairwise disjoint.
$S_{n_1}\times \cdots \times S_{n_k}$ acts on $X^{(1)}\cup\cdots\cup X^{(k)}$.
Define $\tau_i = (1^{(i)}\;2^{(i)})$ and
$$
\sigma_i = \begin{cases}
(1^{(i)}\;2^{(i)}\;\cdots\;n_i^{(i)})& \textrm{if $n_i$ is odd,}\\
(2^{(i)}\;3^{(i)}\;\cdots\;n_i^{(i)})& \textrm{if $n_i$ is even.}
\end{cases}
$$
Here if $n_i$ is odd, then the cycle $\sigma_i$ has odd length $n_i$,
while if $n_i$ is even, then $\sigma_i$ has odd length $n_i-1$.
Whatever the parity of $n_i$, the cycle $\sigma_i$ has odd length.
Everybody knows that $S_{n_i}$ is generated by $\{\tau_i, \sigma_i\}$.
A $k$-element
generating set for $S_{n_1}\times \cdots \times S_{n_k}$ is
$$
G:=\{\tau_1\cdot\sigma_2, \tau_2\cdot\sigma_3,\ldots, \tau_k\cdot\sigma_1\}.
$$
If $\alpha = \tau_i\cdot\sigma_{i+1}$, then
$\alpha^{|\sigma_{i+1}|} = \tau_i$ and $\alpha^{|\sigma_{i+1}|+1} = \sigma_{i+1}$,
so any subgroup containing $\tau_i\cdot\sigma_{i+1}$ contains both
$\tau_i$ and $\sigma_{i+1}$. Hence the subgroup generated
by $G$ contains $\{\tau_i,\sigma_i\}$ for all $i$, and therefore
is the full product.
