It's possible to argue fairly elementarily that Liouville's number
$$ L = \sum_{k=1}^\infty 10^{-k!} $$
is transcendental, by showing directly that for every integer polynomial
$$p(x) = a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$
there is a decimal position in $p(L)$ that must be nonzero.
The powers of $L$ have the form
$$ L^k = c_1 10^{-b_1} + c_2 10^{-b_2} + c_3 10^{-b_3} + \cdots $$
where each of the (all different) $b_i$ is a number that can be written as the sum of $k$ (not necessarily different) factorials, and $c_i$ is some integer $\le k!$ that depends on whether some of the factorials are different.
Now let $n$ be the degree of $p$. Among the $b_i$s we find numbers of the form $B_{h,n}=(h+1)!+(h+2)!+\cdots (h+n)!$, and by choosing $h$ large enough, these numbers can be arbitrarily far from anything that can be written as a sum of fewer than $n$ factorials (which are the ones the appear in the expansion of lower powers of $L$).
So if we choose $h$ large enough, we find a $B_{h,n}$ such that the only terms in $p(L)$ that contribute to the digits around decimal position $B_{h,n}$ is the product of $a_n$ with $C_{h,n}$, which is nonzero.
(Everything to the right of this is the sum of products of some $10^{-b_j}$ with a factor that is at most $n!a_n+(n-1)!a_{n-1}+\cdots+a_1$. The bound of the factor depends only on $p$, so if only we choose $h$ such that that the first $b_j$ after $B_{h,n}$ is separated by at least the length of this bound their contribution cannot reach position $B_{h,n}$.
(Similarly, we can arrange for there to be enough space to the left of $B_{h,n}$ to make room for all of $a_nC_{h,n}$ before the previous $b_j$).