Maximal ideals in $K[X_1,\dots,X_n]$

Hint: Define

$$f:K[X_1,...,X_n]\to K\;\;,\;\;f(g(X_1,...,X_n)):=g(a_1,...,a_n)$$

1) Show $\,f\,$ is a surjective ring homomorphism

2) Use now the first isomorphism theorem for rings

3) Remember: if $\,R\,$ is a commutative unitary ring, an ideal $\,I\leq R\,$ is maximal iff $\,R/I\,$ is a field.

Let $P(X_1, \ldots, X_n)$ a polynomial. Substitute $X_i\mapsto X_i + a_i$ and get $$P(X_1+ a_1, \ldots, X_n + a_n) = \sum c_{\alpha} X_1^{\alpha_1} \ldots X_n^{\alpha_n}$$ and so

$$P(X_1, \ldots, X_n) = \sum c_{\alpha} (X_1-a_1)^{\alpha_1} \ldots (X_n-a_n)^{\alpha_n}$$

Note that $c_{(0,\ldots, 0)} = P(a_1, \ldots, a_n)$. Moreover, $$P(X_1, \ldots, X_n) = c_{(0,\ldots, 0)}+ \sum (X_i - a_i) g_i(X_1, \ldots, X_n)$$ as all the other terms $c_{\alpha}(X-a)^{\alpha}$ are divisible by some $(X_i - a_i)$. Therefore $$P(X_1, \ldots, X_n) - P(a_1, \ldots, a_n) \in (X_1-a_1, \ldots, X_n - a_n)$$

and therefore $P(a_1, \ldots, a_n) \in (P, (X_1 - a_1) , \ldots, (X_n - a_n))$. Assume moreover that $P \not \in (X_1- a_1, \ldots X_n - a_n)$. Then $P(a_1, \ldots, a_n) \ne 0$ and we conclude that $1 = P(a_1, \ldots, a_n)^{-1} \cdot P(a_1, \ldots, a_n) \in (P, (X_1 - a_1), \ldots, X_n - a_n)$. Therefore $(X_1-a_1, \ldots, X_n - a_n)$ is maximal.

Hint $\ \ (I,f) = (I,f\ mod\ I) = (I,f(\bar a))\,\ [\,= 1 \iff f(\bar a)\ne 0\iff f\not\in I]$

Remark $\ $ It is instructive to compare this internal approach to the structural approach mentioned by DonAntonio.