In exercise [1, p. 34 exercise 2.3] we are asked to prove that the normalized DFT matrix given in [1, p. 21, (2.22)] is unitary.
Solution: The normalized N\times N DFT matrix is given by:
\mathbf{W}=\frac{1}{{\sqrt N }}\left[ {\begin{array}{*{20}c}
   1 & 1 &  \cdots  & 1  \\
   1 & {e^{( - j\frac{{2\pi }}{N})} } &  \cdots  & {e^{( - j\frac{{2\pi }}{N} \cdot (N - 1))} }  \\
   1 & {e^{( - j\frac{{2\pi }}{N} \cdot 2)} } & \cdots  & {e^{( - j\frac{{2\pi }}{N} \cdot 2 \cdot (N - 1))} }  \\
    \vdots  &  \vdots  & \vdots  &  \vdots   \\
   1 & {e^{( - j\frac{{2\pi }}{N} \cdot (N - 1))} } & \cdots  & {e^{( - j\frac{{2\pi }}{N} \cdot (N - 1)^2 )} }  \\
\end{array}} \right] (1)


\mathbf{W} \mathbf{W}^H=\frac{1}{N}\left[ {\begin{array}{*{20}c}
   1 & 1 &  \cdots  & 1  \\
   1 & {e^{( - j\frac{{2\pi }}{N})} } &  \cdots  & {e^{( - j\frac{{2\pi }}{N} \cdot (N - 1))} }  \\
   1 & {e^{( - j\frac{{2\pi }}{N} \cdot 2)} } & \cdots  & {e^{( - j\frac{{2\pi }}{N} \cdot 2 \cdot (N - 1))} }  \\
    \vdots  &  \vdots  &  \vdots  &  \vdots   \\
   1 & {e^{( - j\frac{{2\pi }}{N} \cdot (N - 1))} } & \cdots  & {e^{( - j\frac{{2\pi }}{N} \cdot (N - 1)^2 )} }  \\
\end{array}} \right]


\cdot \left[ {\begin{array}{*{20}c}
   1 & 1 & \cdots  & 1  \\
   1 & {e^{(j\frac{{2\pi }}{N})} } & \cdots  & {e^{(j\frac{{2\pi }}{N} \cdot (N - 1))} }  \\
   1 & {e^{(j\frac{{2\pi }}{N} \cdot 2)} } & \cdots  & {e^{(j\frac{{2\pi }}{N} \cdot 2 \cdot (N - 1))} }  \\
    \vdots  &  \vdots  & \vdots  &  \vdots   \\
   1 & {e^{(j\frac{{2\pi }}{N} \cdot (N - 1))} } & \cdots  & {e^{(j\frac{{2\pi }}{N} \cdot (N - 1)^2 )} }  \\
\end{array}} \right] (2)

The above multiplication results in:
\mathbf{W}\mathbf{W}^H =\frac{1}{N} \cdot


\cdot \left[ {\begin{array}{*{20}c}
   N & {\sum\limits_{k = 0}^{N - 1} {e^{(j\frac{{2\pi }}{N}k)} } } &  \cdots  & {\sum\limits_{k = 0}^{N - 1} {e^{(j\frac{{2\pi }}{N} \cdot (N - 1) \cdot k)} } }  \\
   {\sum\limits_{k = 0}^{N - 1} {e^{ - (j\frac{{2\pi }}{N}k)} } } & N & \cdots  & {\sum\limits_{k = 0}^{N - 1} {e^{(j\frac{{2\pi }}{N} \cdot (N - 2) \cdot k)} } }  \\
   {\sum\limits_{k = 0}^{N - 1} {e^{ - (j\frac{{2\pi }}{N} \cdot 2 \cdot k)} } } & {\sum\limits_{k = 0}^{N - 1} {e^{ - (j\frac{{2\pi }}{N}k)} } } & \cdots  & {\sum\limits_{k = 0}^{N - 1} {e^{(j\frac{{2\pi }}{N} \cdot (N - 3) \cdot k)} } }  \\
    \vdots  &  \vdots  &  \vdots  &  \vdots   \\
   {\sum\limits_{k = 0}^{N - 1} {e^{ - (j\frac{{2\pi }}{N} \cdot (N - 1) \cdot k)} } } & {\sum\limits_{k = 0}^{N - 1} {e^{ - (j\frac{{2\pi }}{N} \cdot N \cdot k)} } } &  \cdots  & N  \\
\end{array}} \right] (3)

The elements of the matrix \mathbf{W}\mathbf{W}^H=[w_{ml}] for m \neq l are sums of geometric series which are given by [2, p. 16] S=1+a+a^2+....+a^{(N-1)}=\frac{1-a^N}{1-a}. Thus e.g. for w_{12} the following result is derived:
w_{12}=\sum\limits\limits_{k = 0}^{N - 1} {e^{(j\frac{{2\pi }}{N}k)} }
=\frac{1-{e^{(j\frac{{2\pi }}{N}N)} }}{1-{e^{(j\frac{{2\pi }}{N})} }}
=0.

Similar the other terms w_{ml}, m \neq l of the matrix are also equal to zero and therefore \mathbf{W}\mathbf{W}^H=\mathbf{I}, where \mathbf{I} is the identity matrix. Thus the normalized DFT matrix is unitary. QED.

[1] Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”, Prentice Hall, ISBN: 0-13-598582-X.
[2] Bronstein and Semdjajew and Musiol and Muehlig: “Taschenbuch der Mathematik”, Verlag Harri Deutsch Thun und Frankfurt am Main, ISBN: 3-8171-2003-6.