Upper bound on $n(1-x)^n$ in terms of $n$ and $x$ My specific problem:
For fixed $x\in(0,1),$ I would like to know how large $n$ has to be in terms of $x$ so that $n(1-x)^{n-2}\leq \tfrac{1}{5}.$  Since $\displaystyle\lim_{n\rightarrow\infty}n(1-x)^{n-2}=0$ this must eventually be true. 
I think that this can be accomplished using some upper bound on $n(1-x)^n$ but I haven't found any that are tight enough.  For example, the bound $(1-x)^n\leq \frac{1}{1+nx}$ won't help here.  Thanks in advance for your help!
 A: Set $y=1-x$ and solve $ny^{n-2}=\frac{1}{5}\,$ for $n.$ A solution in terms of LambertW is
$$n = \frac{W_0(\frac{1}{5}y^2\ln y)}{\ln y}.$$
Example $x=0.75$, then $n > 0.01272241834262.$ Check: with $n_1=0.0127\,$ you get
$n_1(1-x)^{n_1-2} \approx 0.19965\,$ and with $n_2=0.0128\,$ the numbers are
$n_2(1-x)^{n_2-2} \approx 0.20120$
Edit: The other real branch $W_{-1}$ will give an upper bound for $n$ with the value $n\approx 4.1953616\,$ for the example $x=0.75.\,$ Check: with $n_3=4.195$ you have $n_3(1-x)^{n_3-2} \approx 0.200083\,$ and with $n_4==4.196$ you get $n_4(1-x)^{n_4-2} \approx 0.199853$
A: First of all, there is really no reason to have the $1-x$, you can do everything again with $n y^n$, since $f(x)=1-x$ is a bijection from $(0,1)$ to itself.
Now to have $y^n < \delta$, it is enough to have $n>\frac{\ln(\delta)}{\ln(y)}$. You need $\delta=\varepsilon/n$. So you in principle need $n>\frac{\ln(\varepsilon) - \ln(n)}{\ln(y)}$. The tight solution to this involves the Lambert W function.
However, we can give an answer which is not tight in terms of elementary functions. As a side lemma, prove that $n^{1/2}>\ln(n)$. (This isn't hard: the minimum of $n^{1/2}-\ln(n)$ occurs at $n=4$, and so it is enough to prove $2>\ln(4)$, which is the same as proving $e>2$.) Then by replacing the $\ln(n)$ with $n^{1/2}$, it is enough to take $n$ larger than all solutions to
$$x+\frac{x^{1/2}}{\ln(y)}-\frac{\ln(\varepsilon)}{\ln(y)}=0.$$
The largest solution is
$$x=\left ( \frac{1}{2} \left ( -\frac{1}{\ln(y)} + \left ( \frac{1}{\ln(y)^2}+\frac{4\ln(\varepsilon)}{\ln(y)} \right )^{1/2} \right )\right )^2.$$
Note that everything is sensibly defined provided $0<\varepsilon,y<1$. For a sanity check, if $\varepsilon$ is so small that the middle term is negligible, then we get essentially $\frac{\ln(\varepsilon)}{\ln(y)}$ as expected. (I made an error that failed this check earlier.)
