Rotation of spin operator itself Consider for example the rotation of the $x$ component of spin by $\pi$ about the $z$ axis. This flips the spin giving $$e^{i\pi S_z}S_xe^{-i\pi S_z}=-S_x.$$
However, can this be proved directly using the commutation relation $[S_z,S_x]=iS_y$ and expanding the exponential?
As $[S_x,[S_z,S_x]]\neq 0$, I cannot use the usual formula to simplify it, but I am hoping there might be a similar formula which requires something like $[S_z,[S_x,[S_z,S_x]]]=0$.
 A: As you know, $\vec S= \vec \sigma /2$, so that
$$
e^{i\pi \sigma_z/2} = i\sigma_z, ~~\leadsto \\
\sigma_z   \sigma_ x   \sigma_z  =- \sigma_x    .
$$
This holds for all representations, since the spin 1/2 irrep is faithful, and the above is a group-theoretic identity, adjoint action. This is to say that, since the object is in the Lie algebra (see below), the combinatorics in all representations will be strictly identical to the above spin 1/2 case, and need not be carried out explicitly!
If you insist on summing the entire series in the standard Hadamard lemma,
$$
e^A B e^{-A}= B + [A,B]+ [A,[A,B]]/2! + ...
$$
the recursively nested commutators are also straightforward to compute.
A: If we consider the function
$$
f(\lambda)  = e^{\lambda A} B e^{- \lambda A},
$$
and consider the Taylor expansion of $f(\lambda)$ around $\lambda = 0$, then the observation that $f(0) = B$, and
$$
\frac{df}{d\lambda} = [A,f(\lambda)], \\
\frac{d^2f}{d\lambda^2} = \left[ A, \frac{df}{d\lambda} \right] = [A,[A,f(\lambda)]],
$$
etc., we find
$$
e^{\lambda A} B e^{- \lambda A} = B + \frac{\lambda}{1!} [A,B] + \frac{\lambda^2}{2!} [A,[A,B]] + \frac{\lambda^3}{3!} [A,[A,[A,B]]] + \cdots.
$$
Now consider the special case where
$$
[A,[A,B]] = \beta B,
$$
which is true for the problem you are interested in.
This results in a simplification where all terms collapse into terms proportional to either $B$ or $[A,B]$. Explicitly,
$$
e^{\lambda A} B e^{- \lambda A} = B + \frac{\lambda}{1!} [A,B] + \frac{\lambda^2}{2!} \beta B + \frac{\lambda^3}{3!} \beta [A,B] + \frac{\lambda^4}{4!}  \beta^2 B + \cdots \\
= B \left\{ 1 + \frac{(\lambda \sqrt{\beta})^2}{2!} + \frac{(\lambda \sqrt{\beta})^4}{4!} + \cdots \right\} + \frac{[A,B]}{\sqrt{\beta}} \left\{ \frac{\lambda \sqrt{\beta}}{1!} + \frac{(\lambda \sqrt{\beta})^3}{3!} + \cdots \right\}.
$$
Then you can compare this to the Taylor series for the hyperbolic functions, obtaining
$$
e^{\lambda A} B e^{- \lambda A} = B \cosh (\lambda \sqrt{\beta}) + \frac{[A,B]}{\sqrt{\beta}}\sinh (\lambda \sqrt{\beta}).
$$
In your case, $A = S_z$, $B = S_x$, $\lambda = i \pi$, and $\beta = 1$. By just plugging into the above formula, you quickly find
$$
e^{i \pi S_z} S_x e^{-i \pi S_z} = -S_x.
$$
