= Solution
The factor $1+e^{-i\xi}$ has a simple zero at $\pi$, and $L(\pi)\ne0$ makes the zero of $m$ there exactly order $N$. The <quadrature mirror filter> construction gives a high-pass symbol $m_1(\xi)=e^{-i\xi}\overline{m(\xi+\pi)}$, up to a constant phase, and
$$
\widehat\psi(\xi)=m_1(\xi/2)\widehat\varphi(\xi/2).
$$
The normalized <scaling function> has $\widehat\varphi(0)=1$, so $\widehat\psi$ has a zero of exactly order $N$ at zero. <Compact support> of the <Daubechies wavelet> permits differentiation under its <Fourier transform>:
$$
\widehat\psi^{(r)}(0)=(-i)^r\int x^r\psi(x)\,dx.
$$
Hence
$$
\boxed{\int x^r\psi(x)\,dx=0\ (0\le r<N),\qquad \int x^N\psi(x)\,dx\ne0.}
$$
There are exactly $N$ <vanishing moments>, rather than just at least $N$.
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