Let $A$ be a commutative ring. I'm trying to prove that in $A[x]$, the Jacobson radical $\mathcal{J}$ is a subset of the nilradical $\mathcal{P}$. This is an exercise from Atiyah & MacDonald
Let $a_0+a_1x+a_2x^2+\dots +a_nx^n\in\mathcal{J}$. Then $$1-(a_0+a_1x+a_2x^2+\dots+ a_nx^n)y,$$ is a unit for all $y\in A[x]$. Putting $y=1$, we get that $$(1-a_0)-a_1x-\dots-a_nx^n,$$ is a unit. This is only possible if $(1-a_0)$ is a unit and $a_i$ are nilpotent for $i\geq 1$.
If I can somehow prove that
$$(1-a_0)\text{ is a unit} \implies a_0\text{ is nilpotent}$$ then I'll be done. This is because if all coefficients $a_0,a_1,\dots, a_n$ are nilpotent, then the polynomial $a_0+a_1x+a_2x^2+\dots +a_nx^n$ too will be nilpotent. However, I'm having problems proving this.
Is my assertion even true? A preliminary investigation shows that it may be false, but I'm not sure. If it is, any helpful hints as to how to prove it would be great.
HINT: In stead of putting $y=1$, try putting $y=x$.