Since you mentioned Krull's height theorem (= the generalized principal ideal theorem) and having difficulty applying it, I thought you or someone else might appreciate seeing how this works: it is quite straightforward.
The generalized principal ideal theorem is as follows: let $R$ be a Noetherian ring and $I$ a proper ideal of $R$ which can be generated by $n$ elements. Let $\mathfrak{p}$ be a prime ideal which is minimal among all primes containing $I$. Then $\mathfrak{p}$ has height at most $n$, that is, there do not exist prime ideals $\mathfrak{p}_0,\ldots,\mathfrak{p}_n$ such that
$\mathfrak{p}_0 \subsetneq \mathfrak{p_1} \subsetneq \ldots \subsetneq \mathfrak{p_n} \subsetneq \mathfrak{p}$.
[For a deduction of this from Krull's Principal Ideal Theorem, see e.g. Theorem 96 on p. 70 of Commutative algebra.]
Let us apply this with $R = k[x_1,\ldots,x_n]$ and the ideal $I = \langle x_1 - a_1,\ldots,x_n - a_n \rangle$. $I$ is itself a maximal -- hence prime -- ideal, since $R/I \cong k$. Thus the generalized principal ideal theorem simply says that $I$ cannot be generated by fewer elements than its height. But its height is certainly at least $n$. No geometry is needed here: just define $\mathfrak{p}_0 = 0$ and for $1 \leq i \leq n$, $\mathfrak{p}_i = \langle x_1 - a_1,\ldots,x_i - a_i \rangle$.
Finally, a comment: I did not use that $k$ was algebraically closed per se but only worked with maximal ideals of this particular form. On the other hand, it is still true over an arbitrary field $k$ that every maximal ideal of $k[x_1,\ldots,x_n]$ has height $n$ and can be generated by $n$ elements (and no fewer, by Krull's theorem): see Corollary 130 on p. 83 of the document linked to above.
