A misconception needs to be cleared up here. Recall that $f(x) = O(g(x))$ means that there exists a constant $C$ such that $|f(x)| \le C |g(x)|$ in a neighborhood of whatever value of $x$ you're interested in. It is an abuse of notation that we use the equality symbol here at all; $O(g(x))$ is not a function, so it cannot be "equal" to $f(x)$ in the usual sense. See the discussion at the Wikipedia article.
Based on this definition, people sometimes say that $f(x) = h(x) + O(g(x))$ if $f(x) - h(x) = O(g(x))$. This is an additional abuse of notation, but it generally does not create problems and can be extremely convenient. I guess you could write $f(x) < h(x) + O(g(x))$ if you really wanted to, but this is redundant, since the inequality is already built into the definition of big-O notation.
But I have no idea what $f(x) > h(x) + O(g(x))$ is supposed to mean. If you are trying to say that there is a constant $C$ such that $|f(x) - h(x)| > C |g(x)|$, the correct way to write this is using big-Omega, not big-O, as $f(x) = h(x) + \Omega(g(x))$.
In a sense, because a factorial is "locally exponential".
What does this mean? Factorials satisfy the recurrence relation
$$(n + 1)! = (n + 1) \cdot n!$$
while exponentials satisfy
$$b^{n + 1} = b \cdot b^n$$
for a given base $b$. Now suppose we have an $n$ that's pretty large. Then $n + 1 \approx n$. Then we have that
$$(n + 1)! \approx n \cdot n!$$
which is similar to the base-$n$ exponential's recurrence, i.e.
$$n^{n+1} = n \cdot n^n$$
In particular, we can even go out some number $k$ of increments in the same way, provided that $k$ is not too big, to get:
$$(n + k)! \approx n^k \cdot n!$$
Hence, over a range of $k$ not so large that any of them are comparable to $n$, $k \mapsto (n + k)!$ is approximately an exponential function! And thus if you plot $y = (n + x)!$ with a large $n$ and not too large range of $x$, you will get an approximately exponential curve, which then becomes likewise approximately linear in the logarithmic plot.
Best Answer
It is enough to take the limit $\frac{n^{\alpha}}{\log n}$ with $\alpha >0$ to see that it tends to infinity and hence we get a stronger relation: $n^{\alpha} = \omega (\log n)$. In other words, the ratio is not bounded above by any constant.