Prove that $n^n$ is greater than $1\cdot3\cdot5\cdots(2n-1)$ 
Prove that $\ n^n \ge (1)(3)(5)\cdots(2n-1)$    

I can't think of how to start answering this question and it would be great help if someone could explain how I should go about doing it.  
Note:It is a sequence and series question(AP,GP,HP)
 A: HINT: If $n=2m$ is even, you can pair up the factors on the righthand side: $1\cdot(2n-1)$, $3\cdot(2n-3)$, and so on until you get to $(2m-1)(2m+1)$. Each of these pairs has the form $(2m-k)(2m+k)$ for some odd $k$ and is therefore less than $(2m)^2=n^2$.

Example. Take $n=4$, so that we’re comparing $4^4$ with $1\cdot3\cdot5\cdot7$. The righthand side is $$(1\cdot7)\cdot(3\cdot5)=\big((4-3)(4+3)\big)\big((4-1)(4+1)\big)<4^2\cdot4^2=4^4\;.$$

If $n$ is odd, you can pair up all of the factors on the righthand side except the one in the middle, which is $n$ itself. Throw that unpaired factor away and compare what’s left with $n^{n-1}$.
A: $$1\cdot3\cdot5\cdot\cdots\cdot(2n-1)=\frac{1\cdot2\cdot3\cdot\cdots\cdot(2n)}{2\cdot4\cdot\cdots\cdot(2n)}=\frac{(2n)!}{2^nn!}$$ 
So you are asking to prove $$(2n)^nn!\geq(2n)!$$
Divide by $n!$ and you are trying to prove 
$$(2n)^n\geq(n+1)(n+2)\cdots(2n)$$
which is obviously true because each side is a product of $n$ factors and the factors on the left are all larger than the factors on the right (except for one factor that they equal).
A: Only noticed this question today. Although the selected answer is quite nice and arguably simpler than the argument below, none of the posted answers address what appeared to be the original intent of establishing the inequality using the Arithmetic Mean-Geometric Mean Inequality. For this, simply notice that
 $$ 1+3+\ldots+(2n-1)=n^2, $$
which can be easily proved, for instance, using induction. By the AM-GM inequality, it follows that 
 $$n=\frac{1+3+\ldots+(2n-1)}n\ge\sqrt[n]{1\cdot 3\cdot\ldots\cdot (2n-1)}, $$
and we are done.
A: Because $\log$ is concave, we have
$$
\frac{1}{n}\log 1+\frac{1}{n}\log 3+\cdots+\frac{1}{n}\log(2n-1)\leq \log\left(\frac{1}{n}(1+\cdots+(2n-1))\right)=\log n.
$$
It remains to multiply both sides by $n$ and exponentiate.
A: In this answer, we present an approach to prove that $n^n\ge (2n-1)!!$ that relies on induction.  Before, proceeding, we provide the following primer.

PRIMER:
In THIS ANSWER, I showed using Bernoulli's Inequality, that the sequence of functions $e_n(x)=\left(1+\frac xn\right)^n$ monotonically increases for all $x>-n$.  
Letting $x=1$, we see that the sequence $$\bbox[5px,border:2px solid #C0A000]{e_n=\left(1+\frac1n\right)^n}\tag 1$$ is monotonically increasing.

We begin the inductive proof be establishing a base case.  Note that for $n=1$, $n^n=1\ge (2n-1)!!=1$.  Now, we assume that $n^n\ge (2n-1)!!$ for some $n>1$.  Then, we write
$$\begin{align}
(n+1)^{n+1}&=(n^n)\,(n+1)\left(1+\frac1n\right)^{n}\\\\
&\ge (2n-1)!!\,(n+1)\left(1+\frac1n\right)^{n}\\\\
&=(2n+1)!!\frac{(n+1)\left(1+\frac1n\right)^{n}}{2n+1}\tag 2
\end{align}$$
Now, we use the result from the primer to assert that $\left(1+\frac1n\right)^{n}\ge 2$.  Inasmuch as the term $\frac{n+1}{2n+1}$ monotonically decreases and is greater than $1/2$, then we have 
$$\begin{align}
\frac{(n+1)\left(1+\frac1n\right)^{n}}{2n+1}&\ge \frac12 \times 2\\\\
&=1 \tag 3
\end{align}$$ 
Using $(3)$ into $(2)$ yields the coveted inequality
$$(n+1)^{n+1}\ge (2n+1)!!$$
and we are done!

NOTE:
Note the term $\frac{(n+1)\left(1+\frac1n\right)^{n}}{2n+1}$ is actually greater than $4/3$ and hence the inequality is strict for $n\ge 2$.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\mrm}[1]{\,\mathrm{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{\left.\vphantom{\Large A}\ln\pars{x}\right\vert_{\ x\ >\ 0}
\,\,\,\,\leq\,\,\,\, x - 1}$:

\begin{align}
\sum_{k = 0}^{n - 1}\ln\pars{2n - 2k - 1} & =
n\log\pars{n} +
\sum_{k = 0}^{n - 1}\ln\pars{2 - {2k + 1 \over n}} 
\\[5mm] & \leq
n\log\pars{n}\ +\
\underbrace{\sum_{k = 0}^{n - 1}\pars{1 - {2k + 1 \over n}}}
_{\ds{=\ \color{#f00}{0}}}
\end{align}
