Why is $n \left(1-p^{\frac{1}{n}}\right)$ increasing in $n$? I would need a proof that $n \left(1-p^{\frac{1}{n}}\right)$ is increasing in $n \in \mathbf{N}$ for any $p \in (0,1)$.
Context
I am working on a larger question and this is the last missing piece. But with this I'm a bit out of ideas (I tried the difference between $n+1$ and $n$ and also derivative wrt $n$, but neither gave anything that seemed useful).
I know how to find limit of this with L'Hospital's rule (it is $-\log p$). But I it does not seem to help to prove the monotonicity.
 A: One could also use elementary estimates.
Fix $n$, and set $p = x^{n(n+1)}$. Now the proving the statement is equivalent to
$$
n(1-x^{n+1}) \leq (n+1)(1-x^{n})
$$
which can be written as
$$
nx^{n}(1-x) \leq (1-x^{n}).
$$
Since $x \in (0, 1)$, we may divide by $1-x$ to get
$$
nx^n \leq 1+x+\ldots + x^{n-1}.
$$
But since $x^{n} \leq x^{k}$, for each $k \in \{0, 1, \ldots, n - 1\}$, the statement follows. (And the inequality is even strict.)
A: 0 < p < 1 : let f : x-> x*(1 - $p^\frac{1}{x}$)
$f'(x) = (1-p^\frac{1}{x}) +\frac{1}{x}*ln(p)*p^\frac{1}{x} = (1-p^\frac{1}{x}) -\frac{1}{x}*ln(\frac{1}{p})*p^\frac{1}{x} $ = $1 - [1+\frac{1}{x}*ln(\frac{1}{p})]*p^\frac{1}{x}$
Let $ h : u-> 1 - [1 + u*ln(\frac{1}{p})]*p^u  = f'(\frac{1}{u}) $
$ h'(u) = u*p^u*[ln(p)]^2 >0 $, when  u>0 
h increases, h(0) = 0, so for u >0, h(u)>0  => $f'(x)$ = $h(\frac{1}{x}) > 0 $
So f increases relatively to x, 
n+1 > n => f(n+1) > f(n) => Your sequence increases
A: Take the derivative to $n$:
$$\frac{d}{dn}n(1-p^{\frac{1}{n}})=1-p^{\frac{1}{n}}+\frac{np^{\frac{1}{n}}\log p}{n^2}=1+\left(\frac{1}{n}\log p-1\right)p^{\frac{1}{n}}$$
This will have to be greater than $0$ for every $n\in\mathbb{N},p\in(0,1)$, so
$$1+\left(\frac{1}{n}\log p-1\right)p^{\frac{1}{n}}>0\iff\left(\frac{1}{n}\log p-1\right)p^{\frac{1}{n}}>-1$$
and this holds for all $n$ and for all $p>0$, so the function $n(1-p^{\frac{1}{n}})$ is increasing in $n$
