It's enough to show that there exists infinitely many primes $ p$ such that all the numbers from $ 1$ to $ n$ are quadratic residues modulo $ p$ because order of quadratic residue is $ \frac {p - 1}{2}$. If we denote $ p_1,...,p_k$ all odd primes less or equal to $ n$, then by Chinese remainder theorem and Dirichlet's Theorem there are infinitely many primes $ q$ such that $ q \equiv a^2 \pmod{p_1 p_2...p_k}$ and $ q\equiv 1 \pmod 8$. For all such primes $ q$, by Quadratic reciprocity law we have $ \left(\frac {p_i}{q}\right) = \left(\frac {q}{p_i}\right) = 1$ and $ \left(\frac {2}{q}\right) = 1$ so also $ \left(\frac {m}{q}\right) = 1$ for $ m\le n$ because of multiplicativity of Legendre symbol.