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I know that $\left\{\frac{1}{\sqrt{2\pi}}e^{inx}\right\}_{n \in \mathbb{Z}}$ is a maximal orthonormal set in $L^2(\mathbb{T}) \approx L^2([-\pi,\pi])$.

How can I use this fact to show that $\left\{\frac{1}{(\sqrt{2\pi})^N}e^{ih\cdot x}\right\}_{h \in \mathbb{Z}^N}$ is a maximal orthonormal set in $L^2([-\pi,\pi]^N)$?

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The basis you want to show is a complete orthonormal basis is $$ E_{n_1,n_2,\cdots,n_N}= \frac{1}{(2\pi)^{N/2}}e^{in_1 x_1}e^{in_2 x_2}\cdots e^{in_N x_N}. $$ You can do this by showing that Parseval's identity holds for all $f\in C([-\pi,\pi]^N)$ because this subspace is dense in $L^2([-\pi,\pi]^N)$. Using this subspace makes it easy to deal with Fubini's Theorem: $$ \sum_{n_1}\sum_{n_2}\cdots\sum_{n_N}|\hat{f}(n_1,n_2,\cdots,n_N)|^2 \\ = 2\pi\sum_{n_2}\cdots\sum_{n_N}\int_{-\pi}^{\pi}|\hat{f}(x_1,n_2,\cdots,n_N)|^2 dx_1 \\ = \cdots =(2\pi)^{N}\int_{-\pi}^{\pi}\cdots\int_{-\pi}^{\pi}|f(x_1,x_2,\cdots,x_n)|^2dx_1 dx_2\cdots dx_N. $$ The notation is messy because of having discrete transforms mixed with untransformed variables.

For example, start with the case where $N=2$. The functions $$ \left\{ \frac{1}{2\pi}e^{in_1x}e^{n_2y}\right\}_{n_1,n_2} $$ form an orthonormal subset of $L^2([-\pi,\pi]^2)$. Suppose $f$ is continuous on $[-\pi,\pi]^2$. Then Bessel's inequality gives $$ \sum_{n_1,n_2}\left|\frac{1}{2\pi}\int\int e^{in_1x}e^{in_2y}f(x,y)dxdy\right|^2 \le \int\int |f(x,y)|^2dxdy. $$ So the unordered sum on the left is absolutely convergent. This allows you to sum however you want, which gives $$ \sum_{n_1,n_2}\left|\frac{1}{2\pi}\int\int e^{-in_1x}e^{-in_2y}f(x,y)dxdy\right|^2 \\ = \sum_{n_1}\sum_{n_2}\left|\frac{1}{\sqrt{2\pi}}\int_{-\pi}^{\pi}e^{-in_2y}\left(\frac{1}{\sqrt{2\pi}}\int_{-\pi}^{\pi}f(x,y)e^{-in_1 x}dx\right)dy\right|^2 $$ The function $\frac{1}{\sqrt{2\pi}}\int_{-\pi}^{\pi}f(x,y)dx$ is a continuous function of $y$ because $f$ is assumed to be jointly continuous in $x,y$ on $[-\pi,\pi]^2$. So it is certainly in $L^2[-\pi,\pi]$, which allows you to apply Parseval's theorem for the variable $y$, which gives the above equal to $$ \sum_{n_1}\int_{-\pi}^{\pi}\left|\frac{1}{\sqrt{2\pi}}\int_{-\pi}^{\pi}e^{-in_1 x}f(x,y)dx\right|^2dy \\ = \int_{-\pi}^{\pi}\sum_{n_1}\left|\frac{1}{\sqrt{2\pi}}\int_{-\pi}^{\pi}e^{-in_1 x}f(x,y)dx\right|^2 dy \\ = \int_{-\pi}^{\pi}\int_{-\pi}^{\pi}|f(x,y)|^2dx dy $$

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    To be honest, I don't find this remark particularly helpful. How can I reduce my problem in order to use the result I know about the one dimensional case? – 2017-02-02
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    @Rene : Sorry about that. I slipped into Parseval identity for Fourier transforms. I've rewritten for you using the series. – 2017-02-02
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    Thank you. Could you add some details to the key steps? For example, where do you apply Fubini? why is it true that you can write $\sum_{h \in \mathbb{Z}^N}$ as multiple sums in that way? why can you use the one dimensional Parseval identity component by component? – 2017-02-02
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    @Rene : I added the detail for $[-\pi,\pi]^2$. The higher-dimensional cases follow the same pattern. – 2017-02-02