Representation of polynomials Let $A$ be a ring and $A[x_1, x_2, ... ,x_n] = A[x_1][x_2]...[x_n]$ be the ring of the polynomials of $n$ independent variables over $A$, i.e. every element $f \in A[x_1, x_2, ... x_n]$ is $$f = f_m x_n^m + f_{m-1}x_n^{m-1} + ... + f_0$$

for some $m \in \mathbb Z^{\geq 0}$ and $f_i \in A[x_1, x_2, ..., x_{n-1}]$ for $0 \leq i \leq m$. I'm trying to prove that every such polynomial $f$ is uniquely represented as a finite sum of monomials $x_1^{\alpha_1} x_2^{\alpha_2} ... x_n^{\alpha_n}$. What I have done so far is this: 

First of all let $S = (\mathbb Z ^{\geq 0})^n$ and 

$$G = \bigoplus_{(\alpha_1, \alpha_2, ..., \alpha_n) \in S} A$$

Then $\psi : G \longrightarrow A[x_1, x_2, ... , x_n]$ defined by

$$\psi (\{ a_{(\alpha_1, \alpha_2, ..., \alpha_n)}\} _ {(\alpha_1, \alpha_2, ..., \alpha_n) \in S}) = \sum _{(\alpha_1, \alpha_2, ..., \alpha_n) \in S} a_{(\alpha_1, \alpha_2, ..., \alpha_n)} x_1^{\alpha_1}x_2^{\alpha_2}...x_n^{\alpha_n}$$

is a surjective group homomorphism and all I have to show is that it's kernel is $0$. My idea is to do this by induction over the number of variables $n$. The case when $n = 1$ is easy, but then I don't know how to use the inductive hypothesis. Please help me! Thanks in advance!
Suppose you have $f=f_m x_n^m + f_{m-1}x_n^{m-1} + ... + f_0 \in A[x_1, x_2, ... x_n]$.  By induction on $n$, each $f_i$ can be written uniquely as a sum of monomials in $x_1,\dots,x_{n-1}$.  Define an element $a\in G$ by saying that $a_{(\alpha_1,\dots,\alpha_n)}$ is defined to be $0$ if $\alpha_n&gt;m$ and the coefficent of $x_1^{\alpha_1}\dots x_{n-1}^{\alpha_{n-1}}$ in the unique representation of $f_{\alpha_n}$ as a sum of monomials if $\alpha_n\leq m$.  Then $$\psi(a)=\sum_{(\alpha_1,\dots,\alpha_n)}a_{(\alpha_1,\dots, \alpha_n)} x_1^{\alpha_1}\dots x_n^{\alpha_n}=\sum_{\alpha_n}x^{\alpha_n}\sum_{(\alpha_1,\dots,\alpha_{n-1})}a_{(\alpha_1,\dots, \alpha_n)}x_1^{\alpha_1}\dots x_{n-1}^{\alpha_{n-1}}=\sum_{\alpha_n} x_n^{\alpha_n}f_{\alpha_n}=f.$$

This shows $\psi$ is surjective.  For injectivity, suppose that $a\in A$ is such that $\psi(a)=\sum_{(\alpha_1,\dots,\alpha_n)}a_{(\alpha_1,\dots, \alpha_n)} x_1^{\alpha_1}\dots x_n^{\alpha_n}=0$.  For each $i$, define $$f_i=\sum_{(\alpha_1,\dots,\alpha_{n-1})}a_{(\alpha_1,\dots, \alpha_{n-1},i)}x_1^{\alpha_1}\dots x_{n-1}^{\alpha_{n-1}}.$$
As above, we can write $\psi(a)=\sum_i x_n^i f_i$.  Since $f_i\in A[x_1,\dots,x_{n-1}]$, $\psi(a)=0$ implies each $f_i$ must be $0$.  By induction, this means that for each $i$, $a_{(\alpha_1,\dots,\alpha_{n-1},i)}=0$ for all $(\alpha_1,\dots,\alpha_{n-1})$.  But this just means that $a=0$.
Not much to prove, really, this is a non-problem.

Assume the result true for $n$ and study it for $n+1$. Note that $A[x_1, \dots, x_n, x_{n+1}] = A[x_1, \dots, x_n] [x_{n+1}]$, as you have noted yourself. To assume that $f$ has two distinct representations $S_1$ and $S_2$ as a sum of monomials is equivalent to showing that $S_1 - S_2$ is a representation not identically zero of the $0$ polynomial.

Assume that $0 \in A[x_1, \dots, x_n] [x_{n+1}]$ can be written as $f_m x_{n+1} ^m + \dots + f_0$ with the coefficients $f_0, \dots, f_m \in A[x_1, \dots, x_n]$. Then, equality of two polynomials in any ring means that they must be equal component-wisely, so $f_i = 0 \ \forall i = 0, \dots, m$, so every representation of $0$ as a sum of monomials must be identically zero. Therefore, your conclusion follows.