Prove the truth of $\sqrt[n]{n} - 1 < \frac{1}{\log(n)}$ when $n>1$ part of a proof I'm writing for homework I came across the following inequality and I feel like I don't have the right tools to solve it.
$$ \sqrt[n]{n} - 1 < \frac{1}{\log(n)}, 1<n\in \mathbb{N}$$
I thought to solve it by induction, but every time I'm getting stuck at simplifying the expression.
 A: Disclaimer: this proof may be unnecessarily complicated, I would be happy to read any simplification or other idea. 
We want to prove
$$n^\frac{1}{n} < 1+ \frac{1}{\log(n)}.$$
Let us elevate both terms to the positive power $\log(n)$, this reduces to:
$$e^\frac{\log(n)^2}{n} < \left(1+ \frac{1}{\log(n)} \right)^{\log(n)}.$$
Observe that the right hand side, which is monotone increasing, goes to $e$ for $n\rightarrow \infty$, but the left hand side, which is monotone decreasing for $n \geq 8$, goes to $1$. Since at $n=8$ the inequality holds we are left with only a finite number of cases to check, where the inequality holds.
A: Maybe not the best answer but here are a few thought :
$$
\begin{align}
n^{\frac1n} &< \frac{1}{\ln(n)} + 1 &\overset{(*)}{\Longleftrightarrow} \\
\frac{\ln(n)}{n} &< \ln \left( \frac{1}{\ln(n)} + 1 \right) &\overset{(**)}{\Longleftarrow} \\
\frac{\ln(n)}{n} &< \frac{\frac{1}{\ln(n)}}{\frac{1}{\ln(n)}+1} = \frac{1}{\ln(n) + 1} &\Longleftrightarrow \\
\ln(n)^2 + \ln(n) &< n &\overset{(***)}{\Longleftarrow} \\
2\ln(n)^2 &< n &\Longleftrightarrow \\
n &< \exp\left( \sqrt{n/2} \right) = \sum_{i=0}^\infty \frac{n^{i/2}}{2^{i/2}\ i!} &\overset{(****)}{\Longleftarrow} \\
n &< \frac{n^3}{5760} &\Longleftrightarrow \\
75.89 &\approx \sqrt{5760}<n 
\end{align}
$$
$(*)$ Taking $\ln(\cdot)$
$(**)$ Since $\ln(x) \geq \frac{x-1}{x}$ 
$(***)$ Assuming $n \geq e$
$(****)$ Retaining only the term $i=6$.
Then you "only" need to check that this holds for $n \leq 75$
A: To show
$\sqrt[n]{n} - 1 < \frac{1}{\log(n)}, 1<n\in \mathbb{N}
$,
the inequalities needed are
$\ln(2) < .7$,
$\frac{\ln(x)}{x}
\le \frac1{e}
$
(since $\frac{\ln(x)}{x}$
has its maximum at $e$),
$\ln(x) \le x-1
$
for
$x \ge 1$
(compare derivatives),
and
$e^x \le \frac1{1-x}$
for $0 < x < 1$
(compare coefficients in 
the power series).
We have
$n^{1/n}
= e^{\ln(n)/n}
\le \frac1{1-\ln(n)/n}
$,
so we are done if
we can show that
$ \frac1{1-\ln(n)/n}
\le 1+\frac{1}{\ln(n)}
$.
This is equivalent to
$1 \le 1+\frac{1}{\ln(n)}-\frac{\ln(n)}{n}-\frac1{n}
$
or
$\frac{\ln(n)+1}{n}
\le \frac{1}{\ln(n)}
$
or
$\ln^2(n)+\ln(n) \le n$.
We can ask Wolfy,
who says that
$f(x)
=x-\ln^2(x)-\ln(x)
$
has a root at
$x=0=0.290734083070702...$
and is positive for
$x > x_0$.
To prove it,
I will show that
$f(x) > 0$ for $x \ge 4$;
computing $f(1), f(2),$
and
$f(3)$ will complete the proof.
$f(1) = 0,
f(2) = 0.826...,
$
and
$f(3) = 0.694...$.
$f'(x)
= \frac{x - 2 \ln(x) - 1}{x}
$
so we are done if
$x - 2 \ln(x) - 1
\ge 0$.
If
$g(x) = x - 2 \ln(x) - 1$,
then
$g'(x) = 1-\frac{2}{x}
\gt 0$
for $x \gt 2$
and
$g(4)
=3-4\ln(2)
\gt 3-4\cdot .7
\gt 0
$
Therefore
$g(x) > 0$
for
$x \ge 4$.
Going back to $f(x)$,
$f'(x) \gt 0$
for $x \ge 4$
and
$\begin{array}\\
f(4)
&=4-\ln^2(4)-\ln(4)\\
&=4-2\ln(2)(2\ln(2)+1)\\
&\gt 4-1.4\cdot 2.4\\
&=0.34\\
&> 0\\
\end{array}
$
A: Consider $g(y) = e^y - y - y^2$. $g'(y) = e^y - 1-2y$, $\ g''(y) = e^y - 2 > 0 $ for $y > \log 2$ and $g''(y) < 0$ for $y < \log 2$. That is, $g'$ is strictly increasing for $y >  \log 2$ and strictly decreasing for $y < \log 2$.
Now, $g'(0) = 0 $, so $g' < 0 $ in $(0,y_0)$ and $g' > 0 $ in $(y_0,\infty)$ for some $y_0 > \log 2$, and $g'(y_0) = 0$. $y_0 > \log 2$ because $g'(\log 2) < 0$.
So, $g$ attains its minimum value in $(0,\infty)$ at $y_0$. 
Also, $y_0 \in (1,3/2)$, since $g'(1) = e-3 < 0$ and $g'(3/2) = e^{3/2} - 4 >0$. We also have $e^{y_0}=1+2y_0$. Then, $g(y_0) = 1+2y_0 - y_0 - y_0^2 = 1+y_0-y_0^2 > 0$, since $1+y-y^2 > 0$ in $(1,(\sqrt{5}+1)/2) \supset (1,3/2) \ni y_0$. Thus $g(y) \geq g(y_0) > 0$ for $y > 0$.
Now, by the Mean Value Theorem, we have $\log(1+1/y) = \log (y+1) - \log(y) = 1/\eta$ for $\eta \in (y,y+1)$. So, $1/\eta > 1/(y+1) > y/e^y$, since $g > 0$.
Therefore, for $y > 0$, $$\log(1+1/y) > \frac{y}{e^y}$$ 
Putting $y = \log x$, we have $\log(1+1/\log x) > \log x/x$, that is,
$$1+\frac{1}{\log x} > x^{1/x} \text{ for } x >1.$$
