If I show that
$$\sum 2^{n\alpha} \sin\left(\frac{x}{\alpha}\right)$$
converges uniformly by the M-test, for all x in $R$, then we may differentiate term-by-term everywhere.
Can I conclude from here that the power series must define a $C^{\infty}$ function?
Or must I show that the differentiated series is also uniformly convergent and hence a continuous function, and then use an induction argument?
1) This is not a power series.
2) It is not enough to show that the series converges uniformly with each summand being differentiable.
3) It is enough to show that the series of derivatives converges uniformly to show that the limit function is differentiable.
More precisely, if each $f_n : I \to \Bbb{R}$ is $C^1$, if $\sum_n f_n(x)$ converges for at least one $x\in I$ and if $\sum_n f_n'$ converges (locally) uniformly, then the series $\sum_n f_n$ converges (locally) uniformly on $I$ and defines a $C^1$ function.
4) You can then in principle use an induction argument to show that the limit is $C^\infty $.
5) In your case, the series is very strange, since the sine term apparently does not depend on $n $ and can thus be pulled out of the series.