= Solution
For an <algebraically closed field> $k$, the <Weak Hilbert Nullstellensatz> says that every maximal ideal of $k[x_1,\ldots,x_n]$ is
$$
(x_1-a_1,\ldots,x_n-a_n)
$$
for a unique $a\in k^n$. Equivalently, every proper ideal has a common zero. The <Strong Hilbert Nullstellensatz> says that for every ideal $I$,
$$
\boxed{I(V(I))=\sqrt I}.
$$
To deduce the strong form, let $f$ vanish on $V(I)$ and introduce a variable $y$. The equations in $I$ together with $1-yf$ have no common zero: a common zero would satisfy both $f=0$ and $yf=1$. The weak theorem therefore gives
$$
1=\sum_i a_i(x,y)g_i(x)+b(x,y)(1-yf(x)),
\qquad g_i\in I.
$$
Substitute $y=f^{-1}$ in the localization $k[x_1,\ldots,x_n,f^{-1}]$. The last term vanishes, and clearing a power of $f$ yields $f^N\in I$. Thus $f\in\sqrt I$. The reverse inclusion is immediate, completing the <Rabinowitsch trick> proof.
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