Proving this inequality $\sqrt[n]{x^n+\sqrt[n]{(2x)^n+\sqrt[n]{(3x)^n+\cdots}}}< (x+\frac{1}{n-1})$ How can I prove this inequality $$\sqrt[n]{x^n+\sqrt[n]{(2x)^n+\sqrt[n]{(3x)^n+\cdots}}}< \left(x+\frac{1}{n-1}\right)$$ if $n$ and $x$  are  positive integer number $$x>=1$$ $$n>1$$

As A.D. noted, the inequality is wrong.  So what is the correct upper bound?
Let
$$ u_{km} = \dfrac{1}{kx}\sqrt[n]{(k x)^n + \sqrt[n]{((k+1) x)^n + \sqrt[n]{\ldots + \sqrt[n]{(mx)^n}}}}$$
with $x \ge 0$, $m \ge k \ge 1$, $n \ge 1 $, so the left side of your inequality  is $\lim_{m \to \infty} x u_{1m}$.
We have
$$ u_{km} = \sqrt[n]{1 + \dfrac{k+1}{k^n x^{n-1}} u_{k+1,m}} \le 1 + \dfrac{k+1}{n k^n x^{n-1}} u_{k+1,m}$$
with $u_{mm} = 1$. 
We get
$$ u_{1m} \le \sum_{j = 0}^{m-1} \dfrac{(j+1)!}{(n x^{n-1})^j (j!)^n} < \sum_{j = 0}^{\infty} \dfrac{(j+1)!}{(n x^{n-1})^j (j!)^n} $$
Since $n \ge 2$, $(j+1)!/(j!)^n \le (j+1)!/(j!)^2 = (j+1)/j!$ and $x^{n-1} \ge x$ so
$$ u_{1m} < \sum_{j=0}^\infty \dfrac{j+1}{(n x)^j j!} = \left(1 + \dfrac{1}{n x}\right) e^{1/(n x)}$$
Thus a correct inequality is 
$$ \sqrt[n]{x^n + \sqrt[n]{(2 x)^n + \sqrt[n]{\ldots }}} < \left(x + \dfrac{1}{n}\right) e^{1/(nx)}$$
