one problem in topology (homeomorphic) show that $(X_1 \times ... \times X_{n-1})\times X_n$ is homeomorphic to $X_1 \times ... \times X_{n-1}\times X_n$
It looks self-evident... but I want formal proof, and actually I cannot exactly know difference between $(X_1 \times ... \times X_{n-1})\times X_n$ and $X_1 \times ... \times X_{n-1}\times X_n$.
 A: You are correct that the problem seems almost tautological.  To simplify, let $n=3$ and let the spaces be called $X$, $Y$, and $Z$ respectively.  There is an “obvious” map $$f \colon (X \times Y) \times Z,\qquad ((x,y),z) \mapsto (x,y,z)$$
and this map is “clearly” a bijection (Scare quotes mean you should verify it if you don't agree that it's clear/obvious).  So $(X \times Y) \times Z$ and $X \times Y \times Z$ are isomorphic as sets, and, one might argue, canonically so.
It remains to be shown that $f$ and $f^{-1}$ are continuous.
The topology on $X \times Y \times Z$ is generated by sets of the form $U \times V \times W$, where $U$ is open in $X$, $V$ is open in $Y$, and $W$ is open in $Z$.  The topology on $(X \times Y) \times Z$ is generated by sets of the form $A \times W$, where $A$ is open in $X \times Y$ and $W$ is open in $Z$.   
To show $f$ is continuous, it suffices to show $f^{-1}(U \times V \times W)$ is open in $(X \times Y)\times Z$ for any open sets $U \subset X$, $V \subset Y$, and $W \subset Z$.  In fact, we claim $f^{-1}(U \times V \times W) = (U \times V)\times W$.  This is but a definition chase:
$$\begin{split}
f^{-1}(U \times V \times W) &= \left\{((x,y),z)\mid (x,y,z) \in U \times V \times W\right\}\\
&= \left\{((x,y),z)\mid x\in U,\ y\in V,\ z\in W\right\}\\
&= \left\{((x,y),z)\mid (x,y)\in U\times V,\ z\in W\right\}\\
&= (U \times V) \times W
\end{split}$$
Since $U$ is open in $X$ and $V$ is open in $Y$, $U\times V$ is open in $X \times Y$.  Hence $(U\times V)\times W$ is open in $(X \times Y)\times Z$.
I will leave the proof that $f^{-1}$ is continuous to you, but it's similar.
A: To condense notation let $A = X_1 \times ... \times X_{n-1}$. We want to show $(A) \times X_n \cong A \times X_n$, although I am inclined to say that $(A) = A$. If you accept this then we have an exercise in showing that a set is homeomorphic to itself. As such, there should be nothing to prove, although to be thorough you could mention that the linear map suffices as a bijection. If you cannot assume $(A) = A$ then try user135988's suggestion in the comments.
