Is the space of immersions of $S^n$ into $\mathbb R^{n+1}$ simply connected?

I'm not a professional topologist by any means, but let me give this a shot. There's some discussion in the first few lectures of John Francis from this course. Please correct me if I've made mistakes below...

By Smale and Hirsch, the space of immersions of $S^n$ into $R^{n+1}$ is homotopy equivalent to the space of unbased maps from $S^n$ into $V_{n}(\mathbb{R}^{n+1})$, $Map(S^n,V_n(\mathbb{R}^{n+1})$. Here $V_n(\mathbb{R}^{n+1})$ is the Stiefel manifold of $n$ frames in $n+1$-space. $V_{n}(\mathbb{R}^{n+1})$ is homeomorphic to $SO(n+1)$.

See Ryan's answer for some more details.

Your question is then basically equivalent to characterizing $\pi_1(Map(S^{n},SO(n+1)))$, the fundamental group of that mapping space.

As B.S. points out in his answer, since $SO(n+1)$ is a group, $Map(S^n,SO(n+1))$ is homotopy equivalent to $SO(n+1)\times\Omega^n(SO(n+1))$, therefore the above fundamental group is $\pi_1(SO(n+1))\times\pi_1(\Omega^n(SO(n+1)))\cong\pi_1(SO(n+1))\times\pi_{n+1}(SO(n+1))$.

Though B.S. has pointed out that we can see that it's always nontrivial just from the first factor, we can in fact compute the group from known results on homotopy groups of $SO(n+1)$ (see e.g. this table compiled from the literature by Klaus Johannson).

The result for all $n$ is (scroll right in the grey box below):

n| 1   | 2      | 3              | 4         | 5    | 6      | 8s-1                | 8s             | 8s+1        | 8s+2      | 8s+3           | 8s+4      | 8s+5      | 8s+6   |
-| --- | ------ | -------------- | --------- | ---- | ------ | ------------------- | -------------- | ----------- | --------- | -------------- | --------- | --------- | ------ |
 | Z   | Z_2+Z  | Z_2+Z_2+Z_2    | Z_2+Z_2   | Z_2  | Z_2+Z  | Z_2+Z_2+Z_2+Z_2     | Z_2+Z_2+Z_2    | Z_2+Z+Z_2   | Z_2+Z_2   | Z_2+Z_2+Z_2    | Z_2+Z_2   | Z_2+Z_4   | Z_2+Z  |

In the rightmost columns, $s$ is any integer greater than or equal to 1, and the plus signs denote direct sum. Apologies for the ugly formatting of the table.

Your space is never simply connected (for $n\geq 1$). As already answered, it is (weakly) homotopy equivalent by Smale-Hirsch h-principle to the space of unbased maps from $S^n$ to $SO(n+1)$, which is itself $SO(n+1)\times \Omega^n SO(n+1)$ ($\Omega^n$ = based maps from $S^n$). So $\pi_1$ is at least $\pi_1(SO(n+1))$ (and more in general). In fact it is not even connected as soon as $\pi_n SO(n+1)$ isn't trivial, which occurs for $n=1,3,4,5$ and many more (maybe all except $n=2$?). But it is connected for $n=2$, which awarded its celebrity to Smale (sphere eversion), even if his advisor Raoul Bott didn't believed it at first. EDIT: according to j.c. $\pi_n SO(n+1)$ is trivial only for $n=2,6$. Quite an interesting fact !

Let me just fill-in the gap in j.c.'s exposition. Smale-Hirsch states that the derivative from the space of immersions $Imm(S^n, \mathbb R^{n+1})$ to the space of bundle monomorphisms $Mono(TS^n, T\mathbb R^{n+1})$ is a homotopy-equivalence.

There is a cute observation that allows you to nail-down the space of fibrewise one-to-one maps $TS^n \to T\mathbb R^{n+1}$.

Notice that $TS^n$ is virtually trivial, i.e. $TS^n \oplus \epsilon^1 = S^n \times \mathbb R^{n+1}$, where $\epsilon^1$ is a line bundle over $S^n$.

So the idea is to extend any bundle monomorphism to an orientation-preserving bundle monomorphism $S^n \times \mathbb R^{n+1} \to T \mathbb R^{n+1}$. You can do this continuously, over the entire space using the a cross-product type construction. This is a homotopy-equivalence between $Mono(TS^n, T\mathbb R^{n+1})$ and the $n$-fold free loop space of $SO_{n+1}$.