Note the Taylor expansion for $\sin{x}$: $$ \sin{x} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots $$
Now consider the Taylor expansion for $\sin(bt)$: $$ \begin{align} \sin{(bt)} &= (bt) - \frac{(bt)^3}{3!} + \frac{(bt)^5}{5!} - \frac{(bt)^7}{7!} + \cdots \\ &= b\left[b^0t - \frac{b^2t^3}{3!} + \frac{b^4t^5}{5!} - \frac{b^6t^7}{7!} + \cdots\right] \end{align} $$
This is where it gets interesting; define $c^{m+1}\equiv b^m$ and substitute it for $b$ in the brackets, $$ \begin{align} \sin{(bt)} &= b\left[ct - \frac{c^3t^3}{3!} + \frac{c^5t^5}{5!} - \frac{c^7t^7}{7!} + \cdots\right] \\ &= b\sin{(c^1 t)} \\ &= b\sin{(b^{1-1} t)} %\\ &= b\sin{t} \end{align} $$
It has been shown $\sin{(bt)} = b\sin{(t)}$, but this does not make sense... surely there is something wrong with the assumption that there is a $c$ such that $c^{m+1} = b^m$?
If you say $c^{m+1}=b^m$ for some specific $m$, that makes sense (but your series calculation is wrong).
What you need for your calculation to work is $c^{m+1}=b^m$ for every even value of $m$, that is, $$c^1=b^0\quad\hbox{and}\quad c^3=b^2\quad\hbox{and}\quad c^5=b^4\quad \hbox{and}\ldots\ .$$ The only possible solutions are $b=1,c=1$ and $b=-1,c=1$. In both of these cases, it is in fact true that $\sin(bt)=b\sin t$.