fundamental group of $GL^{+}_n(\mathbb{R})$

I would like to know whether the $GL^{+}_n(\mathbb{R})$ the set of all invertible matrices with positive determinant is simply connected or not? I guess it is not simply connected but that is just a guess only, I do not know how to prove that, i.e how to show that its fundamental group is non-trivial.

Well, I can rigorously prove that this is connected and hense path connected as it is Lie Group. Could any one rigorously tell me how to approach this kind problem and solve them from the basic knowledge of fundamental group or some other way? So basically I need some result or tools by which I can compute fundamental groups of all known classical matrix Lie groups.

The Gram-Schmidt process shows that $\text{GL}_n^{+}(\mathbb{R})$ deformation retracts onto $\text{SO}(n)$. There is a natural fiber bundle

$$\text{SO}(n-1) \to \text{SO}(n) \to S^{n-1}$$

given by considering the action of $\text{SO}(n)$ on the unit sphere in $\mathbb{R}^n$, and the corresponding long exact sequence in homotopy shows that $\pi_1(\text{SO}(n)) \cong \pi_1(\text{SO}(3))$ for $n \ge 3$. But $\text{SO}(3) \cong \mathbb{RP}^3$ has fundamental group $\mathbb{Z}/2\mathbb{Z}$ (or more explicitly its double cover is $\text{SU}(2) \cong S^3$, which is simply connected), hence so does $\text{SO}(n)$ for $n \ge 3$, hence so does $\text{GL}_n^{+}(\mathbb{R})$ for $n \ge 3$. The cases $n = 1, 2$ are straightforward.

The corresponding double covers of $\text{SO}(n), n \ge 3$ are the spin groups.