TITLE: How to form a mental image of $\mathbf {ijk}=-1$ in quaternions?
QUESTION [18 upvotes]: $\mathbf {ijk}=-1$ is part of the famously stone-carved formula by Sir William Rowan Hamilton, allowing the multiplication of triplets. There is already an intuition question on the topic of quaternions, and a beautifully illustrated post on 3D stereographic projections of quaternion rotations in 4D in the 3Blue1Brown youtube channel.
The question is very specific about how to picture in a geometric or intuitive way the equality $\mathbf {ijk}=-1$ part in the formula
$$\mathbf {i}^2 = \mathbf {j}^2 = \mathbf {k}^2 =\mathbf {ijk}=-1 $$

Attempt: $\mathbf i$, $\mathbf j$ and $\mathbf k$ act on the left of a quaternion as a pure rotation; hence,
$$\mathbf {ijk}\overset{?}=\mathbf {(ij)k}$$
and $\mathbf {i}$ acts on $\mathbf {j},$ rotating $\mathbf {j}$ into $\mathbf {k}$ in keeping with 3Brown1Blue's Grant Sanderson's "right-hand rule", illustrated with the 3D diagram below, corresponding to the stereographic projection of a 4D hypersphere of quaternions with norm $1$ and $0$ real component, where the $-\infty,$ $-\mathbf i,$ $1,$ $\mathbf i,$ $+\infty$ yellow line is the projection of a circle in 4D running through $-1,$ and centered at $0,$ passing through $-\mathbf i$ and $+\mathbf i;$ while the red-blue $\mathbf j,$ $\mathbf k,$ $-\mathbf j,$ $-\mathbf k$ circle corresponds to the 3D projection of a sphere in the 4D hypersphere passing through $-1,$ and centered at $0,$ and reaching both $\pm \mathbf k$ and $\pm \mathbf j.$ Rotating in the direction of $\mathbf i$ (i.e. multiplying on the left by $\mathbf i$) would amount to sliding the yellow line in the direction of the thumb, while rotating the red-blue circle following the clenching of the rest of the fingers:

and resulting in
$$\begin{align}\mathbf {ijk}&\overset{?}=\mathbf {(ij)k}\\
&=\mathbf {kk}\\
&=\mathbf k^2 =-1
\end{align}$$

REPLY [3 votes]: THIS IS NOT AN ANSWER - Just an appendix / extended comment that doesn't belong to the original post...
@anon's answer contains a "belt trick" to illustrate the $\bf{ijk}=-1.$ Explained in words was rather confusing, but (after his pictures holding his belt included in the comments to his post as a link, I believe the idea can be illustrated as follows:

Here is the part of @anon's answer relevant to the illustration:

Since $\mathbf{v}$ has polar form $\exp(\frac{\pi}{2}\mathbf{v})$, the corresponding rotation is $180$ degrees around $\mathbf{v}$ (in either direction). So, take a belt and hold it straight out in front of you (like nunchuks I guess?). Pay attention to the orientation of the belt buckle, as you (in order) rotate it around the $\mathbf{k}$-, $\mathbf{j}$- and $\mathbf{i}$-axes. The effect is the buckle itself will be back to its original position, but the belt will have a $360^{\circ}$ twist in it, represented by the quaternion $-1$.

Background elements:
Unit imaginary quaternions are of the form
$$\left( 0, a{\bf i} + b{\bf j} + c {\bf k } \right)$$
such that $a^2 + b^2 + c^2 = 1.$
It is proven here that
$$\left( a{\bf i }+ b{\bf j} + c {\bf k} \right)^2=-1$$
allowing their expression as
$$e^{\theta \vec {\bf v}}= \cos \theta + \sin \theta\; \vec {\bf v}$$
A rotation of $180$ degrees around the unit imaginary vector $\vec {\bf v}$ will correspond to
$$e^{\frac \theta 2 \vec {\bf v}}=e^{\frac\pi 2 \vec {\bf v}}$$

Here's a link.