Theorem: Liouville's number: $$L=\sum_{i=0}^\infty 10^{-i!}$$ is transcendental.
Proof: Denote $$L_k=\sum_{i=0}^{k!} 10^{-i!}$$ We have: $$L_1=0.1$$ $$L_2=0.11$$ $$L_3=0.110001$$ $$L_4=0.110001000000000000000001$$
We show that for sufficiently large $k$ powers of $L_k^n$ have the same digits in its decimal expansion as $L^n$. Assuming $L$ will be a solution to some polynomial, we will construct a polynomial with infinitely many solutions, which will give a contradiction.
Lemma: Let $n\in \Bbb{N}.$ Let $L_k^n=0,a_1a_2a_3\ldots a_{s}$ be the decimal expansion of $L_k^n$ and $L^n=0,b_1b_2b_3\ldots b_{s}\ldots$ the decimal expansion of $L^n$. There exists $k\in \Bbb{N}$ such that $\forall i\leq k!:a_i=b_i$.
Proof: We have: $$ L_k^n=(10^{-1}+10^{-2}+10^{-6}+\cdots+10^{-k!})\times$$ $$\times(10^{-1}+10^{-2}+10^{-6}+\cdots+10^{-k!})\times$$ $$\times10^{-1}+10^{-2}+10^{-6}+\cdots+10^{-k!}\times\cdots$$ The last digit, that is $a_s$ will be given by multiplying the $10^{-k!}$ terms $n$ times. That is: $10^{-nk!}$.We want $L^n$ to be very same at $b_s$.With the very same multiplication of $L^n$ term given by multiplying "the next", that is $10^{-(k+1)!}$ with $n-1$ terms $10^{-1}$ bigger. Thus following must hold: $$10^{-nk!}>10^{-(n-1)}\cdot 10^{-(k+1)!}$$ equivalently: $$nk!<n-1+(k+1)k!$$ Let $k=n-1$, then we obtain: $$nk!<k+nk!$$ which is valid and we have shown such $k$ exists.
Let's make a little observation. $$L^2=0.0121002200010000000000002200\ldots$$ $$L_1^2=0.01$$ $$L_2^2=0.0121$$ $$L_3^2=0.012100220001$$ $$L_4^2=0.012100220001000000000000220002$$
The argument is as follows: All the nonzero digits are further and further apart, which means the "next" non-zero digit, after raising the number to $n$-th power, won't play any role in creating "further" digits and thus modifying the previous digits. Thus we will end up having the very same digits, up to $a_s$. For arbitrary big $n$ we can find some $k$ dependent on $n$ to have arbitrarily big "sections" of zeros in $L_k^n$. Even when multiplying $L_k^n$ by arbitrarily big constant, we can still have arbitrarily big "sections" of zeros. Thus if $L_k$ can approximate $L$, then $P(L_k)$ can approximate $P(L)$, where $P$ is a polynomial with rational coefficients.
Now, suppose that $L$ is a solution to polynomial $P(x)=p_0+p_1x+\ldots+...p_nx^n$ and some other polynomial $Q(x)=q_0+q_1x+\ldots+...q_mx^m$. Now compare $P$ and $Q$ when approximating by $L_k$. Denote $n=\max\{\deg P, \deg Q\}$. By the lemma we are able to find $k$, dependent on $n$ such that $P(L_k)=Q(L_k)$ but also $\forall j\in \Bbb{N}, j\geq k$ we have that: $$P(L_j)=Q(L_j)$$ Thus the equation $$P-Q=0$$ has infinitely many solutions. But it can have at most $n$. We conclude that $L$ must be transcendental.
What I would like to know is whether my proof is actually valid and can be taken seriously. I would like to hear someone else's opinion on what could be done better, more rigorous, etc. Actually, I would like to make the proof as rigorous as possible, but still elementary as it is.