No map $\mathbb{C}P^n \rightarrow \mathbb{C}P^m$ inducing a nontrivial map $H^2(\mathbb{C}P^m;\mathbb{Z}) \rightarrow H^2(\mathbb{C}P^n;\mathbb{Z})$

This is part of an exercise from Hatcher's Algebraic Topology text. The exercise is as follows:

Using the cup product structure, show there is no map $\mathbb{R}P^n \rightarrow \mathbb{R}P^m$ inducing a nontrivial map $H^1(\mathbb{R}P^m;\mathbb{Z}_2) \rightarrow H^1(\mathbb{R}P^n;\mathbb{Z}_2)$ if $n > m$. What is the corresponding result for maps $\mathbb{C}P^n \rightarrow \mathbb{C}P^m$?

Here is the argument for maps $\mathbb{R}P^n \rightarrow \mathbb{R}P^m$:

We proceed by contradiction. Recall that $H^{\ast}(\mathbb{R}P^m;\mathbb{Z}_2) \approx \mathbb{Z}_2[\alpha]/(\alpha^{m+1})$ and $H^{\ast}(\mathbb{R}P^n;\mathbb{Z}_2) \approx \mathbb{Z}_2[\beta]/(\beta^{n+1})$, where $\alpha$ is a generator of $H^1(\mathbb{R}P^m;\mathbb{Z}_2)$ and $\beta$ is a generator of $H^1(\mathbb{R}P^n;\mathbb{Z}_2)$. Suppose that there exists a map $h:\mathbb{R}P^n \rightarrow \mathbb{R}P^m$ that induces a nontrivial homomorphism $h^{\ast}:H^1(\mathbb{R}P^m;\mathbb{Z}_2) \rightarrow H^1(\mathbb{R}P^n;\mathbb{Z}_2)$. Then $h^{\ast}(\alpha) = \beta$, giving $0 = h^{\ast}(\alpha^{m+1}) = h^{\ast}(\alpha)^{m+1} = \beta^{m+1} \neq 0$, where the final inequality follows from the fact that $m+1<n+1$ and hence the cup product of $\beta$ with itself $m+1$ times $\beta^{m+1} \in H^{m+1}(\mathbb{R}P^n;\mathbb{Z}_2)$ is a generator (and, hence, non-zero).

Now, all solutions that I've seen simply comment that we can use the same argument to show that there is no map $\mathbb{C}P^n \rightarrow \mathbb{C}P^m$ inducing a nontrivial map $H^2(\mathbb{C}P^m;\mathbb{Z}) \rightarrow H^2(\mathbb{C}P^n;\mathbb{Z})$ if $n > m$. We can use the fact that $H^{\ast}(\mathbb{C}P^n;\mathbb{Z}) \approx \mathbb{Z}[\beta]/(\beta^{n+1})$ where $|\beta| = 2$.

However, if we let $\alpha$ be the generator of $H^2(\mathbb{C}P^m;\mathbb{Z})$ and $\beta$ the generator of $H^2(\mathbb{C}P^n;\mathbb{Z})$, how do we conclude that $\beta^{m+1} \neq 0$? Doesn't $\beta^{m+1}$ live in $H^{2(m+1)}(\mathbb{C}P^n;\mathbb{Z})$, which is $0$ if $2(m+1) \geq n+1$? We're only given here that $n > m$.

Thanks!

The crucial point which is not spelled out explicitly in the comments is that $\Bbb{CP}^n$ is a CW complex of dimension 2n. Its cells correspond to the complex vector spaces $\Bbb C^j$, one for each $0 \le j \le n$, which have real dimension 2j.