= Solution
Under the smoothness hypotheses in the <smooth-mask vanishing-moment criterion>, write any nonzero <integer> $j$ as $j=2^r\ell$, where $r\ge0$ and $\ell$ is odd, possibly negative. Iterating the <scaling refinement equation> exactly $r+1$ times gives
$$
\widehat\varphi(2\pi j+t)=\left[\prod_{n=1}^{r+1}m\left(\frac{2\pi j+t}{2^n}\right)\right]\widehat\varphi\left(\frac{2\pi j+t}{2^{r+1}}\right).
$$
The last <MRA low-pass filter> factor is $m(\pi\ell+t/2^{r+1})$. Its <derivatives> of all orders less than $p$ vanish at $t=0$, because $\ell$ is odd and the <MRA low-pass filter> is $2\pi$-periodic. If the remaining factors are $C^{p-1}$ at the displayed arguments, the <product rule> forces every <derivative> of order less than $p$ of the product to vanish. In particular <compact support> and a finite <MRA low-pass filter> supply all these hypotheses. Therefore the <integer-frequency zeros of a scaling function> are
$$
\boxed{\widehat\varphi^{(k)}(2\pi j)=0\quad(j\ne0,\ 0\le k<p),\quad\text{with the stated regularity}.}
$$
Under only the printed assumptions, the exact same finite refinement argument gives the <Peano zero> $\widehat\varphi(2\pi j+t)=o(|t|^{p-1})$. Indeed, the last <MRA low-pass filter> factor has that estimate, the other <MRA low-pass filter> factors are bounded by one almost everywhere by the <quadrature mirror filter> identity, and the last <Fourier transform> factor is bounded because $\varphi\in L^1$. The estimate initially holds almost everywhere and extends to every $t$ by <continuity> of $\widehat\varphi$. It does not imply arbitrary ordinary higher <derivatives>. In the <lacunary scaling-phase regularity counterexample>, $\widehat\varphi=a\widehat\varphi_0$ is nondifferentiable at <dense> dyadic points next to the isolated integer-frequency zeros of the <smooth> base transform. Thus the same regularity qualification is necessary here.
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