Let $k$ be a fixed positive integer and $\Sigma\_k$ the $k$-th symmetric group. By letting $\Sigma\_k$ permuting an orthonormal basis of a $k$-dimensional Euclidean space, there is a "regular representation" 
$$
r\_k: \Sigma\_k\longrightarrow O(k)
$$
which induces a map between classifying spaces
$$
\rho\_k: B\Sigma\_k\longrightarrow BO(k).
$$
Let $X$ be a **finite** CW-complex and a map
$$
f: X\longrightarrow B\Sigma\_k.
$$
For any positive integer $n$, we produce a map by taking self-product of $f$ for $n$-times
$$
\prod\_n f: X\longrightarrow \prod\_n B\Sigma\_k
$$
and have a composition
$$
g\_n: X\overset{\prod\_n f}{\longrightarrow}\prod\_n B\Sigma\_k\overset{\prod\_n \rho\_k}{\longrightarrow}\prod\_n BO(k)\longrightarrow BO(nk)=G\_{nk}(\mathbb{R}^\infty).
$$
where the last map is induced by the inclusion
$$
\prod\_n O(k)\longrightarrow O(nk).
$$
**Question:** I want to prove that for any $k$, any **finite** CW-complex $X$ and any map $f$, there exists a positive integer $n$ such that $g\_n$ is null-homotopic. Is it true? How to prove?



how to prove the $n$-times self-product of a map is null-homotopic
Yes, this is true.
Your map $B\Sigma\_k\rightarrow BO(k)$ gives rise to a map $B\Sigma\_k\rightarrow BO$, i.e. an element in the reduced K-theory group $\tilde{ko}^0(B\Sigma\_k)$.


Now that whole group is probably not torsion, but the Atiyah-Hirzebruch spectral sequence tells you that, if $K$ is a finite-dimensional $CW$ complex with torsion homology groups, $\tilde{ko}^0(K)$ is torsion.


So if you take $K$ to be a suitably modified skeleton of $B\Sigma\_k$ (see below), it follows that there is $n$ such that the map 
$$K\rightarrow B\Sigma\_k\rightarrow BO(k)\rightarrow BO(nk)$$
is null homotopic, where the last map is the times $n$ map.


But if $X$ is a finite-dimensional $CW$-complex, the map $X\rightarrow B\Sigma\_k$ will factor over $K$ if we choose $K$ to be a sufficiently high-dimensional "skeleton".


I was imprecise above about $K$ being a "suitably modified skeleton", let me clarify this:


If $Y$ is a CW-complex with torsion homology groups, then, for each $n$, there is an $n+1$-dimensional CW complex $K$ with torsion homology groups and a map $K\rightarrow Y$ that induces isomorphisms on homology up to degree $n-1$.


To build that $K$, simply take $\tilde{K}$ to be the $n$-skeleton of $Y$. Now the $n$-th homology of that is not necessarily torsion, but since rationalized homotopy and rational homology agree, we can choose maps from $S^n$ to $\tilde{K}$ such that they are nullhomotopic in $Y$, and their images in $H\_n(\tilde{K})$ generate it rationally. But then form $K$ by attaching cells along those maps. $K$ comes with a map to $Y$, and has torsion homology.


