Prove that the ideal $(X_1-a_1,...,X_n-a_n)$ is maximal in $K[X_1,\dots,X_n]$ 
Let $K$ be a field, and $a_1,\dots,a_n \in K$. Prove that the ideal $$(X_1-a_1,\dots,X_n-a_n)$$ is maximal in $K[X_1,\dots,X_n]$.

I tried proving that the only elements outside the ideal are the invertibles of $K$ (I should still prove that this implies maximality, but it shouldn't be too difficult).
Is there a better strategy, or another stategy?
 A: 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. 
A: 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. 
A: I realize that we should avoid responding to other answers, but when they make false statements there should be a way to correct them.
$\mathbb Q$ is a field.  ${X_1}^2-2$ is a prime element in $\mathbb Q\left[ X_1\right]$ which is a p.i.d so ${X_1}^2-2$ generates a maximal ideal. The converse of this statement is false for the field $\mathbb Q$ or any field that is not algebraically closed. If $K$ is algebraically closed, both the statement of the question and its converse are corollaries of the Hilbert Nullstellensatz.  This basic result in algebraic geometry can be found in texts on algebraic geometry, for example Eisenbud's Commutative Algebra with a view toward algebraic geometry, Springer Graduate Texts in Math, vol 150, pp 34--35. 
A: 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.
A: I think this answer might essentially be the same as @orangeskid's answer, but I'm not sure, so I figured I will post it here. If it is the same or incorrect, please let me know and I will delete it.
Let $I = \langle x_1 - a_1, \dots, x_n - a_n \rangle \subset k[x_1, \dots, x_n]$ for $k$ a field. Consider any ideal $I \subsetneq J$, by the Hilbert Basis theorem $J$ is finitely generated, so we can say that $J=\langle x_1 - a_1, \dots, x_n -a_n, f_1, \dots, f_r\rangle$ for finitely many $1 \le i \le r < \infty$, with $f_i \not\in I, f_i \in k[x_1, \dots, x_n]$ and without loss of generality, all of the $f_i$ are not identically equal to zero.
Case 1: At least one of the $f_i$ is constant.
Without loss of generality, $f_r$ is constant, say $f_r \equiv c \in k, c\not=0$ so $J= \langle x_1 - a_1, \dots, c \rangle = \langle x_1 - a_1, \dots, 1 \rangle = k[x_1, \dots, x_n]$.
Case 2: None of the $f_i$ are constant.
Let's choose one of the $f_i$, which by an abuse of notation we will denote simply by $f_i(x_1, \dots, x_n)$.
Then we have that $$f_i(x_1, \dots, x_n) = f_i^m(x_1, \dots, x_n) + f_i^{m-1}(x_1 -a_1,\dots, x_n - a_n)+ \dots + f_i^1(x_1 -a_1, \dots, x_n -a_n)+ f_i(x_1-a_i,\dots,x_n-a_n) $$ where for all $1 < j \le m, \quad f_i^j(x_1, \dots, x_n) := f_i^{j-1}(x_1,\dots,x_n)-f_i^{j-1}(x_1 - a_1, \dots, x_n-a_n)$, and additionally $f_i^1(x_1, \dots, x_n):=f(x_1,\dots,x_n)-f(x_1-a_1, \dots, x_n-a_n)$, and finally that $m$ is the lowest natural number such that $f_i^m(x_1, \dots, x_n)$ is identically constant, thus for any $k >m$, $f_i^k(x_1, \dots, x_n)\equiv 0$. Such an $m$ is guaranteed to exist, because by construction, $\deg(f_i^1) < \deg(f_i)$ and $\deg(f_i^j) < \deg(f_i^{j-1})$ for all $2 \le j \le m$.
Define $c$ to be the non-zero constant such that $f_i^m(x_1, \dots, x_n) \equiv c $. Now clearly we have that $$ f_i(x_1 -a _1, \dots, x_n -a_n) \in \langle x_1 -a_1, \dots, x_n -a_n \rangle \\ f_i^j(x_1-a_1, \dots, x_n -a_n) \in \langle x_1 - a_1, \dots, x_n - a_n \rangle \quad \forall\ 1\le j \le m-1$$  Therefore we have shown that $f_i(x_1, \dots, x_n) = c + g_i (x_1, \dots, x_n)$ for some $g_i \in \langle x_1 - a_1 , \dots, x_n - a_n \rangle$. Therefore we have that $$ J = \langle x_1 - a_1, \dots, x_n - a_n, \dots, f_i, \dots, f_r \rangle \\  = \langle x_1 - a_1, \dots, x_n - a_n, \dots, c + g_i, \dots, f_r \rangle \\ = \langle x_1 - a_1, \dots, x_n - a_n, \dots, c, \dots, f_r \rangle \\ = \langle x_1 - a_1, \dots, x_n - a_n, \dots, 1, \dots, f_r \rangle = k[x_1, \dots, x_n]$$
Thus, in both cases, $I \subsetneq J \implies J = k[x_1, \dots, x_n]$, so therefore $I$ must be maximal, since clearly $I=\langle x_1 - a_1, \dots, x_n - a_n\rangle$ is proper.
