In an article, it says that "Consider the map $\mathbf{RP}^\infty\times\cdots\times\mathbf{RP}^\infty$(n copies) $\to$ $K(\mathbf{Z}_2,n)$", I think this map is the map related to killing homotopy to construct $K(\mathbf{Z}_2,n)$.
Since we know $H^*(\mathbf{RP}^\infty,\mathbf{Z}_2)=\mathbf{Z}[x]$ and $Sq(x)=x+x^2$ for |x|=1. And $H^*(K(\mathbf{Z}_2,n),\mathbf{Z}_2)=\mathbf{Z}_2[Sq^Il_n]$, where I is admissible and has excess less than n.
"Then the induced map $H^*(K(\mathbf{Z}_2,n),\mathbf{Z}_2)\to H^*(\mathbf{RP}^\infty\times\cdots\times\mathbf{RP}^\infty)$ is a monomorphism in degrees smaller than 2n."
I want to know why it is a monomorphism? In my opinion, I think it is hard to show it through Cartan formula of Steenrod squares concerning a product of n elements of degree 1.
Let $x_1x_2\ldots x_n$ denote the element in $H^{n}(\Bbb R P^\infty \times \dots \times \Bbb R P^\infty; \Bbb Z/ 2\Bbb Z)$ that I am assuming is the image of the generator of $H^n(K(\Bbb Z/2\Bbb Z, n); \Bbb Z/ 2\Bbb Z)$
Instead of trying to compute $\operatorname{Sq}^I(x_1x_2\ldots x_n)$ completely, if you just want to demonstrate a monomorphism consider the leading term of this result under the lexicographic order on the polynomials in $\Bbb Z/ 2\Bbb Z [x_1, \ldots x_n]$. I claim that in the relevant degrees each $\operatorname{Sq}^I$ yields a different leading term.
Here are some examples of leading terms, which should indicate the general case: $$ \begin{align*} \operatorname{Sq}^1(x_1x_2\ldots x_n) &= x_1^2x_2\dots x_n + \ldots\\ \operatorname{Sq}^2(x_1x_2\ldots x_n) &= x_1^2x_2^2x_3\dots x_n + \ldots\\ \operatorname{Sq}^k(x_1x_2\ldots x_n) &= x_1^2\dots x_k^2x_{k+1}\dots x_n + \ldots\\ \operatorname{Sq}^2\operatorname{Sq}^1(x_1x_2\ldots x_n) &= x_1^4x_2\dots x_n + \ldots\\ \operatorname{Sq}^3\operatorname{Sq}^1(x_1x_2\ldots x_n) &= x_1^4x_2^2x_3\dots x_n + \ldots\\ \operatorname{Sq}^k\operatorname{Sq}^1(x_1x_2\ldots x_n) &= x_1^4x_2^2\dots x_{k-1}^2x_l\dots x_n + \ldots\\ \operatorname{Sq}^4\operatorname{Sq}^2(x_1x_2\ldots x_n) &= x_1^4x_2^4x_3\dots x_n + \ldots\\ \operatorname{Sq}^k\operatorname{Sq}^i(x_1x_2\ldots x_n) &= x_1^4\dots x_i^4x_{i+1}^2 \dots x_{k-2i}^2 x_{k-2i+1} \dots x_n + \ldots\\ \end{align*} $$ The general pattern is not too hard to verify using the Cartan formula, and the bookkeeping is much less onerous when you're only focusing on leading terms. (I guess you need to show that the operations respect the monomial order to some degree)