Is the residue field of an algebraically closed valued field algebraically closed?

Is the residue field of an algebraically closed valued field $$K$$ with valuation ring $$A$$, algebraically closed?

If I take a polynomial $$f(x)$$ of $$k_A$$ since $$K$$ is algebraically closed it has a root in $$K$$, say $$b$$, so the valuation of $$f(b)$$ is infinity and consequently the valuation of $$b$$ must be infiinity ...

You want to find a root of $$f$$ where $$\deg f \geq 1$$ is a polynomial in $$k_A$$. It is not a polynomial in $$K$$ so you cannot say that it has a root in $$K$$. However, each coefficient $$c_i$$ of $$f=\sum \limits_{i=0}^d c_i X^i$$ is the residue of some element $$a_i$$ of $$A$$. Let's pick such elements $$a_i$$ and consider $$P=\sum \limits_{i=0}^d a_i X^i$$. This has degree $$\deg f \geq 1$$ so it must have roots in $$K$$. Can you justify that every root of $$P$$ lies in $$A$$? Assuming this, fixing such a root $$b$$, it is easy to see that $$f_A(\overline{b})=\overline{P(b)}=\overline{0}$$ where for $$x \in A$$, I denote the residue of $$x$$ by $$\overline{x}$$. So $$k_A$$ is indeed algebraically closed.