Given two real sequences $\{a_n\}$, $\{b_n\}$, we say that $a_n$ is big Omega of $b_n$ and write $a_n=\Omega(b_n)$ as $n\to\infty$ if for some constant $M>0$, $a_n\geq Mb_n$ for all sufficiently large $n$. When $b_n\geq 0$, the essence of this statement is that $a_n$ grows at least as fast as $b_n$.
Thus, if a sequence $\{a_n\}$ satisfies $a_n=\Omega(C^n)$ for every $C>1$, the sequence $\{a_n\}$ grows at least as fast as every exponential sequence $C^n$. Since we can make the growth rate of $C^n$ arbitrarily large by making $C$ sufficiently large, it feels like any such sequence should grow much faster than any exponential in the sense that $a_n=\omega(C^n)$ as $n\to\infty$, that is, for any $M>0$, $a_n\geq M\cdot C^n$ for all sufficiently large $n$. I've produced an argument for this conclusion, and would appreciate it if it was looked over.
Proof
We first establish that we can rewrite $a_n$ as $e^{b_n}$ for all sufficiently large $n$. This will simplify the argument.
Since $a_n=\Omega(C^n)$ holds for every $C>1$, we can see that for some constant $M> 0$, $a_n\geq M\cdot 2^n$ for all sufficiently large $n$, say $n\geq N$. This implies $a_n$ is strictly positive for $n\geq N$, so we can write $a_n=e^{\ln a_n}$ to deduce that $b_n:=\ln a_n$ ($n\geq N$) gives us the desired form for $a_n$.
Any sequence $C^n$ can be written as $e^{cn}$ for some $c>0$. Thus, since $a_n=\Omega(C^n)$ holds for every $C>1$, we have $e^{b_n}=\Omega(e^{cn})$ for every $c>0$. Fix $c>0$. Then for some constant $M>0$, $e^{b_n}\geq Me^{cn}$ for every sufficiently large $n$, that is,
$$e^{b_n-cn}\geq M\text{ for every sufficiently large }n$$ $$\iff b_n-cn\geq \ln M\text{ for every sufficiently large }n$$ $$\iff \frac{b_n}{cn}-1\geq\frac{\ln M}{cn}\text{ for every sufficiently large }n$$ $$\iff \frac{b_n}{cn}\geq 1+\frac{\ln M}{cn}\text{ for every sufficiently large }n$$
The last of these implies
$$\liminf_{n\to\infty}\frac{b_n}{cn}\geq\limsup_{n\to\infty}\left( 1+\frac{\ln M}{cn}\right)=1+0=1$$
yielding $\liminf_{n\to\infty}b_n/(cn)\geq 1$, or $\liminf_{n\to\infty}b_n/n\geq c$. Since $c>0$ was arbitrary, we must have $\liminf_{n\to\infty}b_n/n=\infty$ and consequently $\lim_{n\to\infty}b_n/n=\infty$. Fixing $c>0$, this implies
$$\lim_{n\to\infty}\frac{b_n}{cn}=\infty$$
Thus, since $\lim_{n\to\infty}cn=\infty$,
$$\lim_{n\to\infty}(b_n-cn)=\lim_{n\to\infty}cn\left(\frac{b_n}{cn}-1\right)=\infty$$ $$\implies \lim_{n\to\infty}\frac{a_n}{e^{cn}}=\lim_{n\to\infty}\frac{e^{b_n}}{e^{cn}}=\lim_{n\to\infty}e^{b_n-cn}=\infty$$
The limit $\lim_{n\to\infty}a_n/e^{cn}=\infty$ is equivalent to $a_n=\omega(e^{cn})$. Since $c>0$ was arbitrary, we immediately get $a_n=\omega(C^n)$ for every $C>1$ after reverting to the $C^n$ notation.
Except a detail mentionned in comment, your proof seems correct, but overcomplicated. Simply write:
Or in $O$ and $o$ notations (more familiar to me):
because $$\frac{O(a_n)}{C^n}=O(a_n)o(1)=o(a_n).$$