Induced maps on homotopy groups by self maps of $\mathbb{CP}^n$ Let $f:\mathbb{S}^2\to \mathbb{S}^2$ with degree $d$.
It is well known that the induced map $$f_\ast:\pi_3(\mathbb{S}^2)=\mathbb{Z}\to \pi_3(\mathbb{S}^2)=\mathbb{Z}$$ is given by multiplication by $d^2$.
But, $\mathbb{S}^2=\mathbb{C}P^1$ and $[\mathbb{C}P^n,\mathbb{C}P^n]=\mathbb{Z}$ for $n\geq 1$ and for any $f:\mathbb{C}P^n\to \mathbb{C}P^n$ determined by $d\in \mathbb{Z}$ the induced map $$f_\sharp: H_{2k}(\mathbb{C}P^n)=\mathbb{Z}\to H_{2k}(\mathbb{C}P^n)=\mathbb{Z}$$ is given by multiplication by $d^k$ for $1\leq k\leq n$.
Question: is it true that the map $$f_\ast : \pi_{2n+1}(\mathbb{C}P^n)=\mathbb{Z}\to \pi_{2n+1}(\mathbb{C}P^n)=\mathbb{Z}$$ is determined by multiplication by $d^{n+1}$?
 A: Here is an elementary, if clunky, proof that $|f_*(1)| = d^{n+1}$. I imagine more care with orientations would ensure the sign.
Write $$R:= H^*(\Bbb{CP}^n;\Bbb Z) = \Bbb Z[x]/(x^{n+1}), \;\;\; |x| = 2.$$ The total space of the oriented circle bundle with Euler class $x$ is $S^{2n+1}$; write $$S := H^*(S^{2n+1};\Bbb Z) = \Bbb Z[y]/(y^2), \;\;\; |y| = 2n+1.$$ If $f: \Bbb{CP}^n \to \Bbb{CP}^n$ is labeled by $d$, then $f^*x = dx$ and because $f^*$ is a ring homomorphism, $f^*x^k = d^k x^k$. 
The total space of the oriented circle bundle with Euler class $dx$ is the $2n+1$-dimensional lens space $L^{2n+1}(d,1)$. Write $$L_d := H^*(L^{2n+1}(d,1); \Bbb Z) = \Bbb Z[\tau, y]/(d\tau, \tau^n, y^2), \;\;\;\; |\tau| = 2, |y| = 2n+1.$$
There is a map of circle bundles $\tilde f: L^{2n+1}(d,1) \to S^{2n+1}$ lifting the map $f$. (There are many; choosing one amounts to an identification of $L^{2n+1}(d,1)$ with $f^*(S^{2n+1})$.) 
The Gysin sequence is natural. All pullback maps are ring homomorphisms; the cup product is a module homomorphism. $\delta$ is only a group homomorphism and decreases degree by $1$. The cup products shown increase degree by $2$. $\require{AMScd}$
\begin{CD}
    R @>\smile e(S)>> R @>\pi_S^*>> S @>\delta_S >>R\\
    @Vf^* VV @Vf^* VV @V\widetilde f^* VV @Vf^* VV\\
    R @>>\smile e(L)> R @> >\pi_L^*> L_d @>>\delta_L > R
\end{CD}
What we know is that $e(S) = x$, so the top left arrow is $x^k \mapsto x^{k+1}$, while $e(L) = dx$, so the bottom left arrow is $x^k \mapsto dx^{k+1}$. (As a sanity check, observe that the left square commutes.) 
$\pi_S^*$ is zero for degree reasons. By exactness, $\pi_L^*x$ is cyclic of order $d$, so must be some unit-multiple of $\tau$; but most importantly, there is nothing in the image of $\pi_L^*$ in degree $2n+1$ (again, degree reasons). Because the cup product maps have kernel precisely equal to $H^{2n}(\Bbb{CP}^n)$, we see that $\delta_S(y) = \pm x^n$ and $\delta_L(y) = \pm x^n$, while $\delta_L(\tau^k) = 0$. Because $f^*(x^n) = d^n x^n$, for the right square to commute, we must have $\widetilde f^*(y) = d^n y$. 
(One could figure out the sign by going through the proof of the Gysin sequence but many of these identifications require orientation conventions you'd have to pay attention to.) 
The map $\widetilde f_*: H_{2n+1}(S^{2n+1};\Bbb Z) \to H_{2n+1}(L^{2n+1}(d,1);\Bbb Z)$ was what we were after in that computation - it's multiplication by $\pm d^n$. 
We now need to compare to the induced map on homology and homotopy groups. The Hurewicz map is natural, and we have a square \begin{CD}\pi_{2n+1} S^{2n+1} @>>> H_{2n+1} S^{2n+1} \\
@V\widetilde f_* VV @V\widetilde f_* VV \\
\pi_{2n+1} L^{2n+1}(d,1) @>>> H_{2n+1} L^{2n+1}(d,1)
\end{CD}
The top map is an isomorphism by the Hurewicz theorem; the rightmost map is multiplication by $d^n$. However, the Hurewicz map for the lens space is multiplication by $d$ (the quotient map from $S^{2n+1}$ to $L^{2n+1}(d,1)$ is a covering map of degree $d$; covering maps with connected total space induce an isomorphism on homotopy groups but multiplication-by-$d$ on top homology). Thus by commutativity, $\widetilde f$ induces multiplication-by-$d^{n+1}$ on homotopy groups.
Finally compare the long exact sequences for a fiber bundle: \begin{CD}\pi_{2n+1} S^1 @>>> \pi_{2n+1} S^{2n+1} @>>> \pi_{2n+1} \Bbb{CP}^n @>>> \pi_{2n} S^1 \\
@VVV @V\widetilde f_* VV @Vf_* VV @VVV \\\
\pi_{2n+1} S^1 @>>> \pi_{2n+1} L^{2n+1}(d,1) @>>> \pi_{2n+1} \Bbb{CP}^n @>>> \pi_{2n} S^1 
\end{CD}
Of course, the outer terms are zero, so the inner horizontal maps are isomorphisms. Because $\widetilde f_*$ is multiplcation-by-$d^{n+1}$ (up to sign depending on the many choices we made in this process and some not-entirely-computed boundary maps), we conclude the same of $f_*$. 
A: This is just to provide a bit of background to Dmitri's elegant answer.
You can do this by induction by recognising that $\mathbb{C}P^n$ is the $n^{th}$ projective plane of the $H$-space $S^1$, and in particular may be constructed using the Hopf construction. This iteratively produces quasi-fibrations $\gamma_n:\ast^nS^1\rightarrow \mathbb{C}P^{n-1}$, starting with $n=1$, where $\ast^nS^1$ is the $n$-fold join of $S^1$, and defines $\mathbb{C}P^n$ as the cofiber
$$\ast^nS^1\xrightarrow{\gamma_n} \mathbb{C}P^{n-1}\rightarrow\mathbb{C}P^n.$$
Note that the degree $d$ self map $d:S^1\rightarrow S^1$ is homotopic to the $d$-fold power map (defined with the Lie product), and since $S^1$ is abelian this map is an $A_{\infty}$-map. It follows that there is an induced map $\underline d_n:\mathbb{C}P^n\rightarrow \mathbb{C}P^n$ between the Hopf constructions at each stage.
In particular, at the $(n+1)^{th}$ stage we have $\ast^{(n+1)}S^1\cong \Sigma^n\bigwedge^{n+1}S^1\cong S^{2n+1}$ and $\ast^{n+1}d\cong \Sigma^n\bigwedge^{n+1}d\simeq d^{n+1}$ sitting in a diagram of cofibrations
$\require{AMScd}$
\begin{CD}
    S^{2n+1}@>\gamma_{n+1}>> \mathbb{C}P^n@>>>\mathbb{C}P^{n+1}\\
    @Vd^{n+1}  V V @VV \underline d_n  V@VV \underline d_{n+1} V\\
    S^{2n+1} @>\gamma_{n+1}>> \mathbb{C}P^n@>>>\mathbb{C}P^{n+1}.
\end{CD}
Technically, it now remains to confirm that $\underline d_n$ is indeed the map you describe, but its Friday and the pub's open, so I'm going to stop typing here.
A: I guess it should be enough to prove this statement for one particular example (for each $d,n$), since homotopy classes of self-maps of $\mathbb CP^n$ are classified by their degree (aren't they?).
Let us identify $S^{2n+1}$ with $|z_1|^2+\ldots+|z_{n+1}|^2=1$.
Consider the map $\phi(n,d):S^{2n+1}\to S^{2n+1}$ which is given by 
$$(z_1,\ldots,z_{n+1})\to (z_1^d,\ldots,z_{n+1}^d)\cdot (\sum_i |z_i|^{2d})^{-\frac{1}{2}}.$$
It is not hard to see that this map is of degree $d^{n+1}$, for example, by counting the number of premiages of any point with all non-zero coordinates on unit $S^{2n+1}$. On the other hand we have the Hopf map $S^{2n+1}\to \mathbb CP^{n}$ and the map $\phi(n,d)$ clearly descends to a degree $d$ map $\mathbb CP^n\to \mathbb CP^n$. Thus, for this particular case the map $\phi(n,d)$ is a "geometric realisation" of the map $\pi_{2n+1}(\mathbb CP^n)\to \pi_{2n+1}(\mathbb CP^n)$. This basically completes the calculation in our example.
