prove that there exist a constant $C>0$ such that $p(n) :=$ number of unorders sets of positive integers whose sum is $n \geq e ^ {(c{\sqrt n})}$ $p(n)$ - the number of unordered sets of positive integers whose sum is n.
I proved that $$p(n) \ge {\max_{1\le k\le n}} {\frac {\binom{n-1}{k-1}} {k!} }$$
now i need to deduce that there is an absolute constant c > 0 for which 
$$p(n) ≥ e ^ {(c{\sqrt n})} $$
would appreciate your help with that.
 A: Your $p(n)$ is the number of partitions of $n$. There is a famous asymptotic formula by Hardy and Ramanujan precising your claim. Maybe you find a simpler proof of your estimate in a combinatorics textbook under the heading "partitions".
A: In "RAMANUJAN
TWELVE LECTURES ON
SUBJECTS SUGGESTED BY HIS LIFE AND WORK"
by G. H. Hardy,
on pages 113-115,
there is a reasonably elementary proof
that there are real positive
$a$ and $b$
such that
$e^{a\sqrt{n}}
\lt p(n)
\lt e^{b\sqrt{n}}
$.
The proof is based on
Euler's partition generating function
$F(x)
=\sum_{n=0}^{\infty} p(n)x^n
=\dfrac1{\prod_{k=1}^{\infty}(1-x^k)}
$.
Using a
Tauberian theorem,
it is then shown that
$\log p(n)
\sim c\sqrt{n}
$
where
$c = \pi\sqrt{\frac23}$.
More precise results
follow.
First,
using Cauchy's theorem
in the form
$p(n)
= \dfrac1{2\pi i} \int_C \dfrac{F(x)}{x^{n+1}}dx
$,
where $C$ is a contour
around the origin,
and the functional equation
$F(x)
=\dfrac{x^{1/24}}{\sqrt{2\pi}}\ln(1/x)\exp(\dfrac{\pi^2}{6\ln(1/x)}F(x_1)
$
where
$\ln(1/x)\ln(1/x_1) = 1$,
it is shown that
$p(n)
=\dfrac1{2\pi\sqrt{2}}\dfrac{d}{dn}\left(\dfrac{e^{c\sqrt{n\lambda_n}}}{\lambda_n}\right)
+O(e^{h\sqrt{n}})
$
where
$\lambda_n
=\sqrt{n-\frac1{24}}
$
and
$h < c$.
This implies that,
and is more precise than,
 that
$p(n)
\sim \dfrac1{4n\sqrt{3}}e^{c\sqrt{n}}
$.
The remainder of the chapter
derives the famous
Hardy-Ramanujan
asymptotic series for
$p(n)$.
