Showing a space is homeomorphic to the Hilbert cube. Let $(X_i,T_i)$ be a countably infinite family of topological spaces each of which is homeomorphic to the Hilbert cube.  Show that $\prod_{i=1}^{\infty}(X_i,T_i) \cong I^{\infty}$.  The question also hints to use the fact that $\prod_{j=1}^{\infty}(\prod_{i=1}^{\infty}(X_{ij},T_{ij})) \cong \prod_{i=1}^{\infty}(X_{i1}, T_{i1})$.
Since $(X_i,T_i) \cong I^{\infty} = \prod_{n=1}^{\infty}(I_n,T'_n)$,
$\prod_{i=1}^{\infty}(X_i,T_i) \cong \prod_{i=1}^{\infty} I^{\infty} = \prod_{k=1}^{\infty} (\prod_{n=1}^{\infty}(I_{nk},T'_{nk}))$.  But from here I'm stuck, unless $\prod_{i=1}^{\infty} I^{\infty} = I^{\infty}?$
 A: It's a matter of shuffling indices. The informal argument: every $X_n$ is "really" a countably infinite product of intervals, and we have countably many of those, so in total we have a countably infinite product of intervals, i.e. the Hilbert cube. But of course, this can be formalised.
First note that there is a bijection $s: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$, with inverse $t$. We can use this to indeed show that $\prod_{i=1}^\infty I^{\infty} = I^{\infty}$, as you suggest.
To do this, send a point $((x_n)_n$, where each $x_n \in I^{\infty}$ under a map $H$ to a point in $I^{\infty}$ whose $i$-th coordinate is exactly $(x_m)_k$, where $t(i) = (m,k)$. Or in formal terms $$H((x_n)_n)_i = (x_{p_1(t(i))})_{p_2(t(i))} $$ where $p_1,p_2 : \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$ are the two projections. 
As a map into a product is continuous iff all the compositions with the projections (form the image onto its factors) are continuous and by definition $\pi_i \circ H = \pi_{p_2(t(i))} \circ \pi_{p_1(t(i))}$, $H$ is continuous.
The inverse map $K$ sends $(x_n) \in I^{\infty}$ to $(y_n)_n$ where the $i$-th coordinate of $y_m$ is just $x_{s(m,i)}$, which means, in terms of projections again, that $\pi_i \circ \pi_m \circ K = \pi_{s(m,i)}$, which implies, if you think about it and use the same criterion twice, that $K$ is also continuous, making $H$ a homeomorphism.
This fills the gap in the argument, and answers the last question in your post. 
