Inequality $C\lceil\log{n}\rceil! \geq n^k$ I've been struggling to prove there exist $C$ for $n, n_{0}, \forall k >0 \in \mathbb{R}$
such that $\forall n > n_{0}$:
\begin{equation}C\lceil\log{n}\rceil! \geq n^k\end{equation}
As you might guess, this is from algorithm analysis. Since $n \in \mathbb{R}$, I've tried using a substitution $n = 2^{x}$ to get rid of the logarithm, but was not able to come up with $C$ anyway. Any help appreciated, but proof method would be great.
 A: I just played with the numbers using common sense:
$logn$ is a much smaller function than $n^k$ so I needed to enlarge the former and reduce the latter, which i've done with a big $c$ and a really small $k$.
What I've come up with is:
$k = \frac{1}{20}$
  $c = 200$
I'll leave you deal with the $n_0$, let me know if you need further help
EDIT DUE TO NEW INSTRUCTION:
You said you need the inequality to apply for a certain C for all k and for all n>$n_0$.
According to what I've learnt, it means that we need to find a C that:
$\lim_{n->\infty}\frac{n^k}{C(logn)!} = 0$
Again, played with numbers:
Take $C = 1$, for every k you see it applies, check the $n_0$ you need and you're done.
A: Let $C=1$. We show that after a while, $\lceil \log n\rceil! \ge n^k$.  
Let $m=\lceil \log n\rceil$. Equivalently, by taking logarithms, we see that we want to show that after a while, $\log (m!)\gt k\log n$.
Note that 
$$\log m!=\log 1+\log 2+\log 3+\cdots +\log m.\tag{1}.$$
The right-hand side of (1) is greater than 
$$\int_1^m \log x\,dx=m\log m-m +1.$$
Since $m \gt \log n$, we obtain
$$\log(m!)\gt (\log n)(\log\log n)-\log n+1\gt (\log n)(\log\log n-1).\tag{2}$$
We need to show that after a while, the right-hand side of (2) is greater than $k\log n$.
That will be the case if $\log\log n-1\gt k$. That gives for $n_0$ the possibly very large number $\exp(\exp(k+1))$.   
Remark: We found a quite crude lower bound for $\log(m!)$. Please see the Stirling Approximation for much sharper results. 
