Prove that $\left( \frac{x+1}{k} \right)^x>x!$ for $kI was watching a demonstration of the relation
$$\left( \frac{x+1}{2} \right)^x>x!$$
for all $x > 1$
And I thought about the biggest possible denominator in the LHS such that it will be eventually bigger than the RHS, that is, for which values of $k$ does
$$\left( \frac{x+1}{k} \right)^x>x!$$
hold for some $n<x$ sufficiently large.
By experimentally testing with WolframAlpha, it seems that if $k>e$ there is no such $n$.
Is there a way to prove it?
 A: Better than the Stirling approximation, this interesting work by Milan Merkle provides
a bilateral approximation to $x!$ which reads
$$
\sqrt {2\,\pi } \;{{\left( {x + n} \right)^{\,\,x + n - 1/2} \;e^{\, - \;\left( {x + n} \right)} }
 \over {x^{\,\overline {\,n\,} } }}\;\; \le \;\;\Gamma (x)\; \le \;\sqrt {2\,\pi }
 \;{{\left( {x + n} \right)^{\,x + n - 1} \sqrt {\,x + n + 1/2} \;e^{\, - \;\left( {x + n} \right)} }
 \over {x^{\,\overline {\,n\,} } }}\quad \quad \left| \matrix{
  \;0 < x \hfill \cr   \;0 \le n\; \to \infty  \hfill \cr}  \right.
$$
where $x^{\,\overline {\,k\,} } $ represent the Rising Factorial.
We can rewrite the above as
$$
\sqrt {{{2\,\pi } \over e}} \;\left( {{{x + n} \over e}} \right)^{\,\,x + n - 1/2} {{\;1}
 \over {x^{\,\overline {\,n\,} } }}\;\; \le \;\;\Gamma (x)\; \le \;\sqrt {{{2\,\pi } \over e}}
 \;\left( {{{x + n} \over e}} \right)^{\,\,x + n - 1/2} {{\;1} \over {x^{\,\overline {\,n\,} } }}
\sqrt {\,1 + {{1/2} \over {x + n}}} \quad \left| \matrix{
  \;0 < x \hfill \cr 
  \;0 \le n\; \to \infty  \hfill \cr}  \right.
$$
i.e.
$$
\sqrt {{{2\,\pi } \over e}} \;\left( {{{x + 1 + n} \over e}} \right)^{\,\,x + n + 1/2}
 {{\;1} \over {\left( {x + 1} \right)^{\,\overline {\,n\,} } }}\;\; \le \;\;x!\;
 \le \;\sqrt {{{2\,\pi } \over e}} \;\left( {{{x + 1 + n} \over e}} \right)^{\,\,x + n + 1/2}
 {{\;1} \over {\left( {x + 1} \right)^{\,\overline {\,n\,} } }}\sqrt {\,1 + {{1/2} \over {x + 1 + n}}}
 \quad \left| \matrix{  \;0 < x \hfill \cr   \;0 \le n\; \to \infty  \hfill \cr}  \right.
$$
and taking  $n=0$
$$
\eqalign{
  & \sqrt {{{2\,\pi } \over e}} \;\left( {{{x + 1} \over e}} \right)^{\,\,x + 1/2} \;\; \le \;\;x!\;
 \le \;\sqrt {{{2\,\pi } \over e}} \;\left( {{{x + 1} \over e}} \right)^{\,\,x + 1/2}
 \sqrt {\,1 + {{1/2} \over {x + 1}}} \quad \left| {\;0 < x} \right.  \cr 
  & 1\;\; \le \;\;{{x!} \over {\sqrt {{{2\,\pi } \over e}} \;\left( {{{x + 1} \over e}} \right)^{\,\,x + 1/2} }}\;
 \le \;\sqrt {\,1 + {{1/2} \over {x + 1}}} \quad \left| {\;0 < x} \right. \cr} 
$$
