In Herstein's book "Noncommutative rings" it is said that it is a simple exercise to show that the Jacobson radical of an algebraic algebra is nil. But I can't see why.
Let $A$ be an algebraic algebra. And take $a\in J(A)$. Since $A$ is algebraic there is a polynomial $f(x)=x^n+\alpha_{1}x^{n-1}+\dots+\alpha_{n-1}x+\alpha_n$ s.t. $f(a)=0$.
Also, $Ma=(0)$ for every irreductible $A$-module $M$, since $a\in J(A)$.
(I don't know if the following is helpful) We also have that for any $b\in A$, $$[f(a),b]= [a^n,b]+\alpha_1[a^{n-1},b]+\dots+\alpha_{n-1}[a,b]=0$$ Commute this with $[a,b]$ and we get $$[[a^n,b],[a,b]]+\alpha_1[[a^{n-1},b],[a,b]]+\dots+\alpha_{n-2}[[a^2,b],[a,b]]=0$$ Commute this with $[[a^2,b],[a,b]]$ and so on, $n$ times, and we get a polynomial involving only commutators of $[a^i,b]$ and none of the coefficients $\alpha_i$.
How does that leads to the fact that $J(A)$ is nil?
You can write $f(x) = x^i g(x)$ for some $i$ such that $g(0) \neq 0$. Note that $g(a) \in g(0) 1 + Aa$ is invertible since $a \in J(A)$. It follows that $0 = f(a) = a^i g(a)$ and multiplying by $g(a)^{-1}$ shows $a^i = 0$.