 Solution

Every derivative of order $n+1$ of $\sin 2x$ has magnitude at most $2^{n+1}$. Pair the symmetric nodes $a=i/n$ and $1-a$. For $0\le x,a\le 1$,If there is an unpaired middle node, its factor has magnitude at most $1/2$. Thus the full nodal product is at most $2^{-(n+1)}$, giving the stronger bound $\Vert f-p_n{\Vert }_{\infty }\le 1/(n+1)!$. In particular this implies the requestedThe bound tends to zero by factorial growth, so $\boxed{p_n\to f\text{ uniformly on }[0,1]}$. This conclusion uses smoothness of this particular function and does not assert uniform convergence of equispaced interpolation for arbitrary continuous data.