Proof of an assertion Show $k^k = n^c$ is such that $k = \frac{{\log}n}{\log{\log}n}$. Note: $\log = \log_2$. To see this, take the log of both sides
$k^k = n^c$ becomes $k{\log}k = c{\log}n$. Now, take $k = c * (\frac{{\log}n}{\log{\log}n})$, for $c = 1,000,000$ and $c = .000001$
$k*{\log}k = 1,000,000 * (\frac{{\log}n}{\log{\log}n}){\log}1,000,000 * (\frac{{\log}n}{\log{\log}n})$ too big
$k*{\log}k = .000001 * (\frac{{\log}n}{\log{\log}n}){\log}.000001 * (\frac{{\log}n}{\log{\log}n})$ too small
Therefore, $k$ has to be somewhere in this vicinity, aka $k = \frac{{\log}n}{\log{\log}n}$. Note that $k = 1 * (\frac{{\log}n}{\log{\log}n})$, when c = 1
Pick $k = \frac{{\log}n}{\log{\log}n}$
$k*{\log}k = \frac{{\log}n}{\log{\log}n}{\log}\frac{{\log}n}{\log{\log}n}$ and somehow, this is supposed to equal $c{\log}n$? I'm unaware of a way to effectively simplify the expression. Tips and tricks or any help with the simplification would be appreciated
 A: Let's take the question and log both sides just as you have done:
$$k \log{k} = c \log{n}$$
Using:
$$k = \frac{\log{n}}{\log{\log{n}}}$$
We get:
$$\frac{\log{n}}{\log{\log{n}}} \log{\frac{\log{n}}{\log{\log{n}}}} = c \log{n}$$
$$\frac{\log{n}}{\log{\log{n}}}\bigg(\log{\log{n}}-\log{\log{\log{n}}}\bigg) = c \log{n}$$
Simplifying:
$$1-\frac{\log{\log{\log{n}}}}{\log{\log{n}}}= c$$
Let's use the following equation to simplify that mess:
$$X = \log{\log{n}}$$
Yielding:
$$1 - \frac{\log{X}}{X} = c$$
$$X^{\frac{1}{X}} = 2^{1-c}$$
Consider $c = \frac{1}{2}$, this gives us:
$$X^{\frac{1}{X}} = 2^{\frac{1}{2}}$$
Therefore, $X = 2$ and since $X = \log{\log{n}}$, $n = 16$
Let's check the equation with $c = \frac{1}{2}$ and $n = 16$
$$k ^ {k} = n^c$$
LHS = $k = \frac{\log{n}}{\log{\log{n}}}$ = 4, RHS = 4. I'm not sure if their intention is for us to derive that functional expression for $k$ but using the expression which they provided to us, the above illustrates how we can use it to determine a value for $n$ and $c$ in the form $n^c$ as the only solution.
A: Exact real solution of the equation
$$k^k = n^c$$
can be expressed via Lambert W-function,
$$k=e^{W(c\ln n)}=\dfrac{c_1\log_2 n}{W(c_1\log_2 n)},$$
where $\;c_1=c\ln 2.\;$
If $\;x\in(0,1000),\;$ then
$$W(x)\approx 0.525\log_2(x).$$

However, for $\;x>>1000\;$ it does not work.

