What does algebraic independence mean? If you search for algebraic independence, you will find the following on Wikipedia:

In abstract algebra, a subset S of a field L is algebraically independent over a subfield K if the elements of S do not satisfy any non-trivial polynomial equation with coefficients in K. 

https://en.wikipedia.org/wiki/Algebraic_independence
What does that mean in simpler terms? I am especially interested in the Brownawell-Waldschmidt theorem (here), but I dont see how it follows that for example at least one of $e\pi$ and $e^{\pi^2}$ is transcendental.
 A: It means that if $n\in\mathbb N$, if $p(x_1,\ldots,x_n)$ is a non-zero polynomial with coefficients in $K$ and is $s_1,\ldots,s_n\in D$, then $p(s_1,\ldots,s_n)\neq0$.
Now, suppose that $e\pi$ and $e^{\pi^2}$ are both algebraic. Then $\{e,\pi\}$ would be algebraically dependent over $\mathbb Q$, since $e\pi$ being algebraic means that there are rational numbers $a_0,\ldots,a_n\in\mathbb Q$, not all of which are $0$, such that$$a_0+a_1e\pi+a_2(e\pi)^2+\cdots+a_n(e\pi)^n=0.$$So, if $p(x,y)=a_0+a_1xy+a_2(xy)^2+\cdots+a_n(xy)^n$, $p(e,\pi)=0$.
A: I'll explain what it means if $S$ is a finite set. Let's write $S=\{a_1,...,a_n\}$. It is algebraically independent if there is no non-zero polynomial $f\in K[x_1,...,x_n]$ such that $f(a_1,...,a_n)=0$. You can try to prove as an easy exercise that this definition is equivalent to the following:
$1.$ Each $a_i$ is transcendental over $K(a_1,...,a_{i-1},a_{i+1},...,a_n)$
$2.$ $a_1$ is transcendental over $K$ and each $a_i$ is transcendental over $K(a_1,...,a_{i-1})$
Now, if $S$ is infinite then it is called algebraically independent if every finite subset is algebraically independent. 
A: Reading the wiki article a little bit further, we can see a very easily comprehensible explanation:
The numbers $x$, $y$ are algebraic independent if there is no 2-variable polynome with rational coefficients for which $P(x,y)$ would be zero.
For example, $\sqrt{\pi}$ and $2\pi+1$ are not algebraically independent, because substituting $x=\sqrt{\pi}$ and $y=2\pi+1$ into the polynome $P(x,y)=2x^2-y+1$ we get zero.
