Prove that T is not a zerodivisor in A[T] 
Let $A$ be any commutative ring. Consider the polynomial ring $A[T]$. Prove that $T$ is not a zerodivisor in $A[T]$. Generalize the argument to prove that a monic polynomial 
  $$
 f=T^n+a_{n-1}T^{n-1}+\dots+a_0
$$

is not a zerodivisor in $A[T]$. 
The question is found in Miles Reid's Undergraduate Commutative Algebra. In an I.D., the proof is simple. I assume this is also the case for an arbitrary ring, but somehow I am unable to find a beautiful=simple proof. Thank you in advance for your contributions. 
 A: HINT: 
Multiplying by $f$ increases the degree by exactly $n$. Look at the term of highest degree in both factors. 
A: Ok, let's try.
Proof: (by contradiction)
Any element $f \in A[T]$ is of the form $f=a_nT^n+a_{n-1}T^{n-1}+\dots+a_0$ , where $a_i \in A$ and $T \in A[T]$.
If $\ T$ is a zerodivisor in $A[T]$ $\implies \exists f \in A[T]$ where $f\neq 0 (\implies a_i\neq 0)$ s.t. $\ fT=0$
$$ fT=a_nT^{n+1}+a_{n-1}T^n+\dots+a_0T
$$
$fT=Tf=0 \implies T=0 \ (\implies A[T]=\{0\})$ or $ a_i=0$ contradicting the assumption that $a_i\neq0$ 
$\therefore T$ is not a zerodivisor  in $A[T]$ $\ \ \blacksquare$
Corrections, comments and alternative proof's are much appreciated. 
A: The idea is that the leading coefficient of the product won't be zero, and the zero polynomial is the one that has all coefficients zero.
The formalism is a bit boring, but we can do it.
Let $A$ be an arbitrary commutative ring with $1$. The set of polynomials in $A$ is defined as follows:
$A[T]:=\{ f: \mathbb{N} \rightarrow A | supp(f) $ is finite $\}$
Support of $f$ is the set of elements of its domain where the value of $f$ is not zero.
$A[T]$ is a commutative ring with $1$ with the following operations:
$(f+g)(n):=f(n)+g(n)$ for all $n\in \mathbb{N}$
$(fg)(n):=\sum_{l+k=n} f(l)g(k)$ for all $n\in \mathbb{N}$
You can check that this definition of polynomial ring works, that the set with thosw operations is a commutative ring with $1$. Also convince yourself that the zero element of the ring is $f(n)=0$ for all $n\in \mathbb{N}$.
Now let $f\in A[T]$ be a monic polynomial of degree $n\in \mathbb{N}$. With our definition, this means that $f(n)=1$ and $f(j)=0$ whenever $j>n$.
Let $g\in A[T]$ be a polynomial of degree $m\in \mathbb{N}$. Again, this means that $g(m)=a\in A\setminus \{ 0 \}$ and $g(j)=0$ whenever $j>n$.
$(fg)(n+m)=1a=a\in A\setminus \{ 0 \}$
Therefore, $fg$ is not the zero polynomial and $f$ is not a zero divisor.
