What is wrong with following trivial argument in proving that fundamental groups of higher order spheres are trivial?
As we know $\mathbb{S^n} - p$ is homeomorphic to $\mathbb{R^n}$, where $p$ is pole of sphere.
So, all loops based on some point of sphere say $s$, not passing through $p$ will be nullhomotopic as $\mathbb{R^n}$ is convex space.
Now apply some rotation about any axis. Given new sphere will be homeomorphic to older one just with transformed pole from $p$ to $p'$ and now all loops based on $s$ passing through $p$ will also be nullhomotopic.
So, all loops based on $s$ are nullhomotopic and so $\pi_1(\mathbb{S^n},s) ={0}$
The problem with this argument is that there are loops $\alpha:I\to \mathbb{S}^n$ which are surjective, they cover entire sphere. Such loops are known as space-filling curves and they exist for any $n\geq 1$ by the Hahn-Mazurkiewicz theorem. Such loops always pass through (any) $p\in \mathbb{S}^n$ regardless of applied rotation (or any other homeomorphism). And so there is no valid choice of $p$ (as a point that $\alpha$ misses).
However it is the case that every loop is homotopic to a non-surjective loop. You can apply the following reasoning Proving directly that $S^2$ is simply connected: is a surjective loop homotopic to a non-surjective one? to any $S^n$ for $n\geq 2$. And with that missing piece your proof is complete.