Radical Ideals: Show that $\sqrt{\sqrt{I}+\sqrt{J}}=\sqrt{I+J}$

The direction $\sqrt{\sqrt{I}+\sqrt{J}}\supset \sqrt{I+J}$ is trivial as $\sqrt{I}+\sqrt{J} \supset I+J$ since $\sqrt{K} \supset K$ for any ideal $K$. Is the following correct?

My attempt: Suppose $f\in \sqrt{\sqrt{I}+\sqrt{J}}$ then $f^n \in \sqrt{I}+\sqrt{J}$ which means that $f^n$ can be written as the sum $g+h$ where $g^p\in I$ and $h^q \in J$. If we raise $f^n$ to the $t=p+q$ then we get $f^{nt}=(g+h)^t=g^t+tg^{t-1}h+\cdots+tgh^{t-1}+h^t$. The first $p$ terms have a factor $g^{p'}$ where $p'\geq p$ which means they are in $I$ and the next (and final) $q$ terms have a factor $h^{q'}$ where $q'\geq q$. Thus $f^{nt}$ is in $I+J$.

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This is roughly the right answer, but you need to correct your expansion of $(g+h)^t$ and pick the right $t$. ($t=2$ isn't gonna work.)
Sorry, it was supposed to say $t=p+q$. That works, right? –  Steven-Owen Feb 25 '13 at 21:40
Yes, except your expansion is wrong. For example $(g+h)^2\neq g^2 + gh + h^2$. –  Thomas Andrews Feb 25 '13 at 21:40