Fourier series and convolution Let $f$ and $g$ be $2\pi$-periodic, piece-wise smooth functions having Fourier series $f(x)=\sum_n\alpha_ne^{inx}$ and $g(x)=\sum_n\beta_ne^{inx}$, and define the convolution of $f$ and $g$ to be $f\star g(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)g(x-t)dt$. Show that the complex form of the Fourier series for $f\star g$ is $$f\star g(x)=\sum_{n=-\infty}^{\infty}\alpha_n\beta_ne^{inx}.$$
I have been trying different approaches to this for a while but I haven't been able to figure it out. I think that I am supposed to use the fact that
$$\frac{1}{2\pi}\langle f,g\rangle=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\overline{g(t)}dt=\sum_{n=-\infty}^{\infty}\alpha_n\overline{\beta_n}$$
but I can't figure out how to manipulate the algebra to show the correct result. I may be forgetting some key property of the inner product or something.

Let the Fourier Series representation for $f$ and $g$ be written respectively by
$$f(x)=\sum_{n=-\infty}^\infty \alpha_ne^{inx}$$
$$g(x)=\sum_{n=-\infty}^\infty \beta_ne^{inx}$$
with coefficients $\alpha_n$ and $\beta_n$ given by 
$$\alpha_n=\frac1{2\pi}\int_{-\pi}^\pi f(t)e^{-int}\,dt$$
$$\beta_n=\frac1{2\pi}\int_{-\pi}^\pi g(t)e^{-int}\,dt$$
The convolution of $f$ and $g$ is defined as
$$\begin{align}
(f*g)(x)&\equiv \frac1{2\pi}\int_{-\pi}^\pi f(t)g(x-t)\,dt\\\\
&=\frac1{2\pi}\int_{-\pi}^\pi f(t)\left(\sum_{n=-\infty}^\infty \beta_n e^{in(x-t)}\right)\,dt\\\\
&=\sum_{n=-\infty}^\infty \beta_n e^{inx}\left(\frac1{2\pi}\,\int_{-\pi}^\pi f(t)e^{-int}\,dt\right)\\\\
&=\sum_{n=-\infty}^\infty \alpha_n\,\beta_n e^{inx}
\end{align}$$
and we are done!
