Prove: if $n = \frac{k^k-1}{k-1}$ then $k=\Omega\left(\frac{\log{n}}{\log{\log{n}}}\right)$ I have to prove that for integers $n$ and $k$, $k\geq2$, and 
$$
n = \frac{k^k-1}{k-1}
$$
follows, that 
$$
k=\Omega\left(\frac{\log{n}}{\log{\log{n}}}\right).
$$
So it would be sufficient to show, that for some constant $c>0$ following inequality holds:
$$
k \geq c\cdot\frac{\log{n}}{\log{\log{n}}}
$$
In detail it would look like:
$$
\frac{\log{n}}{\log{\log{n}}} = \frac{\log{\left(k^k-1\right)}-\log{\left(k-1\right)}}{\log{\left(\log{\left(k^k-1\right)}-\log{\left(k-1\right)}\right)}} \leq \ldots \leq c\cdot k
$$
I'm trying to find an upper bound for the numerator and an lower bound for the denominator of the above fraction, but I haven't worked it out yet. The estimation for the nominator looks e.q. like this:
$$
\log{\left(k^k-1\right)}-\log{\left(k-1\right)} \leq \log{k^k} - \log{\left(k-1\right)} = k\cdot\log{k} - \log{\left(k-1\right)} \leq k^2 - \left(1-\frac{1}{k-1}\right) = k^2 - \frac{k-2}{k-1}
$$
It would be nice if somebody could help me with this assignment!

Geometric series:
$$\frac{k^k-1}{k-1}=1+k+k^2+k^3+\dots+k^{k-1}<k^{k-1}+k^{k-1}+k^{k-1}+\dots+k^{k-1}=k^k$$
Thus,
$$n<k^k$$
$$\log n<\log k^k=k\log k$$
$$k>\frac{\log n}{\log k}$$

if $k\ge3$,
$$1+k+k^2+\dots+k^{k-1}>1+2+2^2+\dots+2^{k-1}>2+2+2+\dots+2=2^k$$
$$n>2^k$$
$$k<\log_2n$$
One can adjust this to see that for any base, the following is eventually true for large enough $k$.
$$k<\log n$$

Combining these, we have
$$k>\frac{\log n}{\log k}>\frac{\log n}{\log\log n}$$
