If $x_1, \dots, x_n$ are elements of $A$, $k[x_1, \dots, x_n]$ usually indicates the subalgebra of $A$ generated by $x_1, \dots, x_n$ (in your case $A$).
Let us write $k[X_1, \dots, X_n]$ for the polynomial algebra on the variables $X_1, \dots, X_n$. Then $A$ is isomorph to a quotient of $k[X_1, \dots, X_n]$, more precisely one can consider the homomorphism $ \begin{array}{c} \varphi: k[X_1, \dots, X_n] \to A\\ p(X_1, \dots, X_n) \mapsto p(x_1, \dots, x_n) \end{array} $
$\varphi$ is surjective by hypothesis and you have $A \cong k[X_1, \dots, X_n]/\mbox{Ker}\,\varphi$.
Algebraic independence is equivalent to the condition $\mbox{Ker}\,\varphi = (0)$. So, in this case, $A$ is isomorph to $k[X_1, \dots, X_n]$.