Is it possible to prove that $3$ is a primitive root of any Fermat prime without quadratic reciprocity?

Hint from Ireland/Rosen which avoids explicitly using quadratic reciprocity:

If $3$ is not a primitive element, show that $3$ is congruent to a square. Use exercise $4$ (which states that if $q$ is a prime and $q \equiv 1 \pmod{4}$, then $x$ is a quadratic residue $\bmod q$ if and only if $-x$ is a quadratic residue $\bmod{q}$) to show that there is a number $a$ such that $-3 \equiv a^2 \pmod{p}$. Now solve $2u \equiv - 1 + a \pmod{p}$ and show that $u$ has order $3$. This would imply that $p \equiv 1 \pmod{3}$, which cannot be true.

Let $n$ denote the number $2^{2^{n'}}+1$, $g$ be a primitive root modulo $n$, and let $n(a)$ denote the order of $a$ modulo $n$, for an integer $a$. Since $a^{n(a)}\equiv g^{n-1} \pmod{n}$, if $n$ is a prime, $a\equiv g^{(n-1)/n(a)} \pmod{n}$; hence $a$ is a quadratic residue modulo $n$ if, and only if, (n-1)/n(a) is even. Take for $n$ the number $2^{2^{n'}}+1$, we see that $a$ is a quadratic residue modulo $n$ if and only if $a$ is not a primitive root modulo $n$.
As $2^{2^{n'}}+1\equiv 2 \pmod{3}$, for every $n'$>1, $n$ is of the form $3x+2$. If 3 is a quadratic residue modulo $n$, however, $n$ must divide a number of the form $x^2-3y^2$, hence $\equiv 1 \pmod{3}$, which is obviously a contradiction. Therefore 3 is not a quadratic residue modulo $n$, thus 3 is a primitive root modulo $n$.
Please forgive me for some typing error.
Thanks in any case for paying attention.