Discrete math question which is as follows:
Prove that $(x+1)!$ is not $O(x!)$ using only the definition of big-O notation. (Hint!: $\log(ab) = (\log a + \log b)$)
I used a proof by contradiction saying that if it was $O(x!)$, by definition: $(x+1)! \leq (c)x!$ for $x > k$ for some $k$. This simplifies to $(x+1)(x!) \leq c(x!)$ where the factorials cancel and we get $x+1 \leq c$ for $x \gt k$ for some $k$. We can always pick an $x$ that breaks this condition no matter what. Therefore $(x+1)!$ is not $O(x!)$.
Is this correct? I did not need to use the $\log(ab)$ property at all, and therefore am skeptical.
Your proof is correct. Note that factorial is particularly amenable to solution using the "limit" definition of big-O:
$$ f(n) = O(g(n))\iff \limsup_{n\to\infty} \frac{|f(n)|}{g(n)} \lt +\infty. $$
Substituting, we get
$$ \limsup_{n\to\infty} \frac{(n+1)!}{n!} = \limsup_{n\to\infty} \,(n+1) = +\infty $$
so $(n+1)!$ is not $O(n!)$.