Proving that $n$-component Brunnian link is nontrivial I stumbled upon the attached image. It shows a way to construct an $n$-component Brunnian link for any $n\geq 3$. That is, this link is not trivial, but deleting any of its components makes the new link trivial. The latter property is obvious from the picture, however I would like to have a strict proof of nontriviality. I have not managed to come up with such a proof, although I do have some kind of a plan.
Let us denote this link by $L=K_1\cup K_2 \cup \dotsb \cup K_{n-1}\cup K_n$. Also, it might be useful to examine $L'=L\setminus K_n=K_1\cup K_2 \cup \dotsb \cup K_{n-1}$.
I look at the link group (i.e., fundamental group of the link's complement) $\pi_1(L').$ Since $L'$ is equivalent to $n-1$ disjoint circles, I conclude that $\pi_1(L')=F_{n-1}$. Then I ask what $[K_n]\in \pi_1(L')$ would be in the case if $L$ were trivial. In that case I would be able to move $K_n$ a little bit and get a homotopic loop that is a clean circle, thus $[K_n]\in \pi_1(L')$ is the identity.
Now, for contradiction I would like to calculate explicit form of $[K_n]\in \pi_1(L')$ (or to somehow just prove that it is not the identity). I tried doing it with Wirtinger presentation, but it was messy and, even if I did find explicit form for $[K_n]\in \pi_1(L')$ in terms of group’s generators, determining whether $[K_n]$ is equal to the identity in this finitely generated group is a well-known undecidable problem.
Is there a way to improve my approach in order to prove nontriviality? If not, do you see another way to prove that?
Thank you.

 A: One thing we can take advantage of is the $n$-fold rotational symmetry about an axis that passes through the center of the link. Let's say $L\subseteq S^3$ is our Brunnian link (or $L\subseteq \mathbb{R}^3$ if you wish). The quotient of $S^3$ by this rotational symmetry is still $S^3$, and the image of $L$ through this quotient is the following knot $K\subseteq S^3$:

By putting the basepoint along the axis of rotation, one can see that the induced map $\pi_1(S^3\setminus L)\to \pi_1(S^3\setminus K)$ is surjective: given a loop in $S^3\setminus K$, lift the path up to one of the $n$ lifts in $S^3\setminus L$, then notice you get a loop. (The quotient map $S^3\setminus L\to S^3\setminus K$, by the way, is an example of a branched cover, where the image of the axis of rotation is a branch/singular locus of order $n$.)
Hence, $L$ is nontrivial if $K$ is nontrivial. One can identify $K$ as a connect sum of two trefoil knots, and hence $L$ is nontrivial.
You might also want to see that $L$ is non-split -- I think this follows from the fact that it's Brunnian.

a well-known undecidable problem

This is true for the general word problem, but the word problem for 3-manifolds is decidable.
