Is following true (and what is the proof)?
$$n|k|^{n-1} \le C\frac1{|k|log\frac1{|k|}}$$ $n$ is a positive integer, $|k|$ is less then $1$, $k$ can be imaginary, $C$ is a real constant.
(The origin is this book)
(Exercise 4.2.3 says just ‘Verify that’ )
We know that for $x=0$ and as $x$ tends to infinity, this expression goes to $0$, so the trick is to find out what is going on in the middle. Calculus is very useful in this regard, so let us consider the derivative of this function.
Expanding with product rule we get $$\frac{d}{dx}(xk^{x-1}) = x\ln(k)k^{x-1} + k^{x-1} = (x\ln(k)+1)k^{x-2}$$ Since $k$ is always positive, we only have to worry about the $x\ln(k)+1$ term (note that $\ln(k)$ is negative). It is easily shown that this linear term is positive for $x<-\frac{1}{\ln(k)}$, $0$ for $x=-\frac{1}{\ln(k)}$, and negative for $x>-\frac{1}{\ln(k)}$. From this we have shown that the function will achieve a maximum at $x = -\frac{1}{\ln(k)} $.
Plugging this in to the original equation, we get $$-\frac{1}{\ln(k)}k^{-\frac{1}{\ln(k)}-1}=\frac{1}{\ln(\frac{1}{k})}\frac{1}{ek}$$ Thus we have your bound with $C=\frac{1}{e}$ $$n|k|^{n-1} \le \frac{1}{e}\frac1{|k|\ln\frac{1}{|k|}}$$