I justed started studying the sorting algorithm, so I need help solving problems on (big Omega) $\Omega$
How can I Prove that $n! = \Omega(n^{100})$
I know that we write $f(x) = \Omega(g(x))$ if $g(x) = O(f(x))$. This means that there is a constant $c>0$ and a value $x_0$ such that $|f(X)| \ge cg(x)$ whenever $x>x_0$.
Hence from the definition above, I can write $$n^{100} = O(n!)$$
We can find a constant c and a value $x_0$ such that $n^{100} \le O(n!)$ for all $x>x_0$. We could take $c=1$ and $x_0=1$
I don't know if I am correct. Please how should I continue and complete the proof.