$\mathbf {Problem:}$ State whether the following claim is True or false : $O(n^{2n})=\Omega(2^n)$. Justify your answer.
$\mathbf {My Attempt:}$ We know that, $f(n) = O(g(n))$ means there are positive constants $c$ and $N$, such that $0 \le f(n) \le cg(n)$ for all $n \ge N$. The values of $c$ and $N$ must be fixed for the function $f$ and must not depend on $n$. Informally, saying some equation $f(n) = O(g(n))$ means it is less than some constant multiple of $g(n)$.
Similarly, $f(n) = Ω (g(n))$ means there are positive constants $c$ and $N$, such that $ f(n) \ge cg(n) $ for all $n \ge N$. The values of $c$ and $N$ must be fixed for the function $f$ and must not depend on $n$.
But I don't have an idea what $O(n^{2n})=\Omega(2^n)$ means. My guess is that , we have to show $\exists$ $c\ge0,N\in\mathbb{N}$ such that $O(n^{2n})\le c\cdot2^n$. But this seems to be vague and unclear.
Please help me understand the problem. Any hint regarding the solution will be highly appreciated.
Strictly speaking, any claim of the form $O(f(n))=\Omega(g(n))$ cannot be true, since on the left side can go any function less than a constant times $f(n)$, and on the right side can go any function greater than a constant times $g(n)$, so by either choosing extremely tiny functions on the left side, or by choosing extremely large functions on the right side, you can contradict it.
In this specific case, it can also be shown that $n^{2n}$ grows faster than any constant times $2^n$, since $$ \frac{n^{2n}}{2^n}=\left(\frac{n^2}{2}\right)^n. $$