I think it's not so trivial to show that $\sqrt{p_n}$ is not already in the field $\mathbb Q(\sqrt{p_1},\ldots,\sqrt{p_{n-1}})$. Bare-hands elementary algebra and unique factorization in $\mathbb Z$ do suffice for $n=1,2,3$, and perhaps a little beyond, but things quickly turn ugly. I do not know what the context is at that point in Isaacs' text, but I can offer two reasonably-structured approaches. The first is easy to explain in words, but is uses some algebraic number theory, which I suspect is not what is intended: in the corresponding rings of algebraic integers, adjoining $\sqrt{p_n}$ introduces ramification at $p_n$. Ignoring the prime $2$, the earlier adjoinings did not introduce ramification at $p_n$... done. The other approach first observes that $\sqrt{p_n}$ is a value of a quadratic Gauss sum, which exhibits that square root as lying in the cyclotomic field obtained by adjoining a $p_n$-th or $4p_n$-th root of unity. The "independence" comes from the "fact" that the degree of the $N$th cyclotomic field over $\mathbb Q$ is Euler totient-function of $N$, and that we have an explicit description of the Galois groups, if needed. Thus, there is a Galois automorphism fixing all those square roots but one, etc. The usual proofs about independence of cyclotomic fields are perhaps not so transparent... and I confess that the only proof I easily remember uses Dirichlet's theorem on primes in arithmetic progressions to obtain a prime $p$ so that all but $\mathbb Q(\zeta_{p_n})$ collapse to give trivial extensions of $\mathbb Q_p$.
EDIT: in response to comment/question... my in-quotes use of "independence" is meant to refer, albeit imprecisely, to the general problem of showing that a given polynomial in $\mathbb Q[x]$ has no root in some field extension of $\mathbb Q$. The exercise as posed asks about "independence" of square roots of primes: $\sqrt{p_n}$ ought not lie in the field extension of $\mathbb Q$ obtained by adjoining square roots of other primes. Similarly, the Gauss-sum argument still does depend on the "independence" of roots of unity of relatively-prime orders. The ramification argument from algebraic number theory is a different device to prove such things.
LATER EDIT: As in Geoff Robinson's answer, there are more elementary arguments! A day in which one learns something so basic is a good day!