He discovered the following result, called Liouville's theorem (or at least this is one result called Liouville's theorem - there's a different one in complex analysis).

Suppose $\alpha$ is algebraic of degree $n$ (that is it is the root of a $n$th degree irreducible polynomial with integer coefficients). Then you can find $c>0$ such that for all rational $p/q$ (with $q>0$) we have $|\alpha -p/q|>c/q^n$.

To prove this, let $P$ be an irreducible polynomial of degree $n$ with integer coefficients such that $P(\alpha )=0$. $P$ has bounded gradient in $[\alpha -1,\alpha +1]$ so there exists $c>0$ such that for any rational $p/q$ in $[\alpha -1,\alpha +1]$ we have:

$|\alpha -p/q|>c|P(\alpha )-P(p/q)|$

(Note that we don't need to worry about rationals outside $[\alpha -1,\alpha +1]$ since these have $|\alpha -p/q|\ge 1$ so can't possibly affect the theorem.)

Now just note that $|q^{n}P(p/q)|$ is a non-negative integer and it is not zero (as $P$ is irreducible) so it is at least 1. And $P(\alpha )=0$. Hence:

$|\alpha -p/q|>c/q^n$

as required.

This theorem allows us to construct many transcendental numbers. The key is to find a number which is so well approximated by rationals that the conclusion of Liouville's theorem cannot possibly hold for any $n$, thus proving the number is transcendental. For example it is an easy exercise now to show that $1/2^{0!}+1/2^{1!}+1/2^{2!}+\dots$is transcendental. Historically this was the first number to be proved as such.

Consider more generally the sum $x=\sum _{i=1}^{\infty }1/a_i$ where $a_i$ are strictly increasing positive integers with $a_i|a_{i+1}$. Suppose this is algebraic of degree $n$. We will use Liouville to derive a necessary condition on the $a_i$s.

We can find $c>0$ such that $|x-p/q|>c/q^n$ for any rational $p/q$. Take $p/q=\sum _{i=1}^{m}1/a_i$. Then $q=a_m$. We get:

$\sum _{i=m+1}^{\infty }1/a_i>c/a_m^n$

and this must hold for all $m$.

Let's estimate the thing on the left hand side. We have $a_i\ge 2^{i-m-1}{a}_{m+1}$ for all $i\ge m+1$. So:

$\sum _{i=m+1}^{\infty }1/(2^{i-m-1}{a}_{m+1})>c/a_m^n$

Using the formula for a geometric progression this gives us:

$2/a_{m+1}>c/a_m^n$ for all $n$.

In other words, if $x$ is algebraic of degree $n$ then there exists $A>0$ such that $a_{m+1}<A{a}_m^n$ for all $m$.

So to ensure that $x$ is not algebraic we just need to make sure this condition is violated for every positive integer $n$. For instance if (for every $n$) $a_{m+1}/a_m^n$ tends to infinity as $m$ tends to infinity, then the condition is certainly violated. If you set $a_m=2^{m!}$ then $a_{m+1}/a_m^n=2^{(m+1)!-n*m!}$ which tends to infinity as $m$ tends to infinity (for any fixed $n$).