In setting up a Fourier series on $\mathbb{R}^n$, we use that for $l, m \in \mathbb{Z}^n$ $$\int_{[0,2\pi]^n}e^{i\langle x, l-m\rangle }=\begin{cases} 0 &\text{ if }l\neq m \\ (2\pi)^n & \text{ if } l =m.\end{cases}$$
The second case is obvious, but I'm missing how to show the first case. Is it simple as in the $n=1$ case?
Yes, the integral can be Fubinized into $$ \prod_{k=1}^n \int_0^{2\pi} e^{ix_i(l_i-m_i)} \,dx_i $$ where $l\ne m$ ensures that at least one of the factors is zero.