I'm trying to prove that an arbitrary single qubit unitary (read: unitary two by two matrix, and thus rotation up to a phase) can be composed from Hadamard and T gates, given by $ H = \frac{1}{\sqrt{2}}\left( \begin{array}{cc} 1 & 1 \\ 1 & -1\\ \end{array} \right)$, $ T = \left( \begin{array}{cc} 1 & 0 \\ 0 & e^{i \pi/4}\\ \end{array} \right)$
The proof is not so difficult, but I'm having some trouble with one of the first steps: showing that $HTH = e^{-i \frac{\pi}{8} \sigma_x}$
I suppose this should follow rather easily, but honestly I don't see it. I know that you can write $T = e^{-i \pi/8 \sigma_z}$ and that $H$ is a pi rotation about the $\frac{1}{\sqrt{2}}(\hat{x}+\hat{z})$ axis, but I don't know how to write that in exponentials, and I don't know how to get the matrix product (which I can work out of course) into the desired exponential form either.
I assume this should be rather trivial, but I just can't find it sadly.
For exponentials involving the Pauli matrices its generally a good idea to use the formula: $$e^{i\theta\sigma}=\cos\left(\theta\sigma\right) + i\sin\left(\theta\sigma\right)$$
Then use the Taylor series for $\cos$ and $\sin$: $$e^{i\theta\sigma} = \sum_{n=0}^\infty \frac{(i\theta\sigma)^{2n}}{(2n)!} + \sum_{n=0}^\infty \frac{(i\theta\sigma)^{2n+1}}{(2n+1)!}$$
Then use that the Paulis are Hermitian and unitary so $\sigma^2 = I$: \begin{align} e^{i\theta\sigma} &= \sum_{n=0}^\infty \frac{(i\theta)^{2n}}{(2n)!} I + \sum_{n=0}^\infty \frac{(i\theta)^{2n+1}}{(2n+1)!}\sigma\\ &= \cos(\theta)I + i\sin(\theta)\sigma \end{align}
Can you see how to apply this technique to your problem?