Second homotopy class of complex Grassmannian Is it true that the second homotopy class of complex grassmannian
$$Gr_{\mathbb{C}}(n, N)= \frac{U(N)}{U(n)\times U(N-n)}$$
is always $\mathbb{Z}$ for arbitrary $N\geq 2$, $1\leq n\leq N-1$, i.e., 
$$\pi_2(Gr_{\mathbb{C}}(n, N))=\mathbb{Z}$$
If it is true, how to prove this? Many thanks!
There are some examples I have found to be true:
When $N=4, n=2$, 
see What are the first 8 homotopy groups of the complex Grassmannian $G_\mathbb{C}(2,4)$?
 A: Yes. The Grassmannian $\text{Gr}(n, n+m)$ fits into a fiber bundle
$$U(n) \times U(m) \to U(n+m) \to \text{Gr}(n, n+m)$$
and applying the long exact sequence in homotopy produces an exact sequence ending
$$\cdots \to \mathbb{Z}^2 \to \mathbb{Z} \to \pi_1(\text{Gr}(n, n+m)) \to 0$$
where we have used that $\pi_1(U(n)) \cong \mathbb{Z}$ for all $n \ge 1$. Now, the map $\mathbb{Z}^2 \to \mathbb{Z}$ is surjective (in fact it is precisely the map $(a, b) \mapsto a + b$), from which it follows by exactness that $\pi_1(\text{Gr}(n, n+m)) = 0$; that is, the complex Grassmannian is simply connected. The exact sequence continues
$$\cdots \to \pi_2(U(n) \times U(m)) \to \pi_2(U(n+m)) \to \pi_2(\text{Gr}(n, n+m)) \to \mathbb{Z}^2 \to \mathbb{Z} \to 0$$
and now we need to know that $\pi_2(U(n)) = 0$ for all $n \ge 1$; there are various ways to prove this, for example using the above long exact sequence for $m = 1$ and inducting on $n$. This gives a short exact sequence
$$0 \to \pi_2(\text{Gr}(n, n+m)) \to \mathbb{Z}^2 \to \mathbb{Z} \to 0$$
which gives $\pi_2(\text{Gr}(n, n+m)) \cong \mathbb{Z}$ as desired.
Alternatively, once you know that the Grassmannian is simply connected, you can use the Hurewicz theorem to reduce the computation of $\pi_2$ to the computation of $H_2$. The homology and cohomology of Grassmannians is completely understood.  
