Let $\mathbb{k}$ be a field and $n\geq 1$. Consider the $\mathbb{k}$-algebra $A$ defined as: $$A:=\mathbb{k}[x]/\langle x^n\rangle.$$ Then the Jacobson radical of $A$ (the intersection of all maximal ideals of $A$) is the ideal $\mathrm{rad}(A)=\big\langle \overline{x}\big\rangle$.
Is the following proof correct ?
Let $f\in\mathbb{k}[x]$. Recall that the Jacobson radical has the following caracterization:
$$\overline{f}\in\mathrm{rad}(A) \Longleftrightarrow \forall g\in\mathbb{k}[x] \ \exists\,h\in\mathbb{k}[x]:\, \overline{(1-fg)h}=\overline{1}.$$
($\Rightarrow$) Suppose that $\overline{f}\in\mathrm{rad}(A)$. Then for each polynomial $g\in\mathbb{k}[x]$ we have: $$\bigg(1-f(x)g(x)\bigg)h(x)=1+x^{n}w(x),\ \text{for some}\ w\in\mathbb{k}[x]$$ Looking at constants temrs we get $$\big(1-f(0)g(0)\big)h(0)=1$$ we must have $f(0)g(0)\neq 1$, for each $g\in\mathbb{k}[x]$. This implies that $f(0)=0$, that is $f(x)$ is a multiple of $x$. Thus for some polynomial $q\in\mathbb{k}[x]$ we have: $$\overline{f(x)}=\overline{xq(x)}=\overline{x}\,\overline{q(x)}\in\langle \overline{x}\rangle.$$
($\Leftarrow$) Conversely, suppose that $\overline{f(x)}\in\langle \overline{x}\rangle$ and for each $g\in\mathbb{k}[x]$ we set $$h=1+fg+\cdots+f^{n-1}g^{n-1}$$ then $$(1-fg)h=1-f^ng^n.$$ Hence $$\overline{f}\in\mathrm{rad}(A) \Longleftrightarrow f^{n}g^{n}\in\langle x^n\rangle,\forall g\in\mathbb{k}[x]$$ but, we know by hypothesis that $\overline{f(x)}\in\langle \overline{x}\rangle$, then $$\overline{f(x)^n}=\overline{f(x)}^{\,n}=\left(\overline{x}\overline{q(x)}\right)^n=\overline{x}^{\,n}\overline{q(x)}^{\,n}=\overline{x^n}\overline{q(x)}=\overline{0}\,\overline{q(x)}=\overline{0},$$ so $\ f^{n}\in\langle{x^n}\rangle$, finallly $$f^{n}g^{n}\in\langle{x^n}\rangle,\ \forall\,g\in\mathbb{k}[x]$$ This implies that $\ \overline{f}\in\mathrm{rad}(A).$