I know there are many ways of showing this isomorphism but I am wondering if this following one works.
Define a ring homomorphism $\mathbb C[x_1,\dots, x_n]\rightarrow \mathbb C$ by $x_i\mapsto a_i$, then this factors through the quotient to give a map $\phi:\mathbb C[x_1,\dots, x_n]/\langle x_1-a_1,\dots, x_n-a_n\rangle$.
Now let $\mathbb C\hookrightarrow \mathbb C[x_1,\dots, x_n]$ be the inclusion, and compose with the proejction $\pi:\mathbb C[x_1,\dots, x_n]\rightarrow \mathbb C[x_1,\dots, x_n]/\langle x_1-a_1,\dots, x_n-a_n\rangle$ to obtain a map $\psi:\mathbb C\rightarrow \mathbb C[x_1,\dots, x_n]/\langle x_1-a_1,\dots, x_n-a_n\rangle$.
I will show that these are inverses of one another. Let $z\in \mathbb C$, then $\psi(x)=[z]\in \mathbb C[x_1,\dots,x_n]/\langle x_1-a_1,\dots, x_n-a_n]$, the map $\phi$, then take $\phi([z])$ clearly equals $z$ so $\phi\circ \psi$ is the identity. Now let $[p]\in \mathbb C[x_1,\dots,x_n]/\langle x_1-a_1,\dots, x_n-a_n\rangle$, then $[p]$ can be written as: $$[p]=\sum_{i_1\cdots i_n}[z_{i_1\cdots i_n}]\cdot [x_1]^{i_1}\cdots [x_n]^{i_n}$$ so: $$\phi([p])=\sum_{i_1\cdots i_n}z_{i_1\cdots i_n}\cdot a_1^{i_1}\cdots a_n^{i_n}$$ applying $\psi$ we obtain that: $$\psi\circ\phi([p])=\sum_{i_1\cdots i_n}[z_{i_1\cdots i_n}]\cdot [a_1^{i_1}]\cdots [a_n^{i_n}]$$ However, in $\mathbb C[x_1,\dots, x_n]/\langle x_1-a_1,\dots, x_n-a_n\rangle$ we have $[a_i^{i_1}]=[x_i]^{i_1}$ implying that $\psi\circ \phi([p])=[p]$ so $\psi\circ \phi$ is the identity and thus both are isomorphic.
In particular I am worried that this is incorrect as most other proofs I have seen rely on polynomial division, or apply Hilbert's Nullstellensatz and Zariski's lemma (which kind of feels like taking a sledgehammer approach), but I guess these also prove that every maximal ideal is of the form above, which my proof does not.
