$2n! \geq (n-k)! n^{k-1} (2n+k-k^2)$ 
Let $0 < k \leq n$ and $k$ is an integer. Then prove that :
  $$2 \cdot (n)! \geq (n-k)! n^{k-1} (2n+k-k^2)$$

The main problem I found in trying to prove this inequality is the term : $2n+k-k^2$. It’s hard to really have a good bound on this polynomial that will help solving this inequality. 
We can notice that we only need to look at the $k$ for which : $2n+k-k^2 > 0$. Yet trying to compute the determinant in order to find the root of the polynomial is a bit cumbersome.
Moreover it’s quite hard to have an analytic way of proving this since taking derivative is impossible here. 
Maybe there is some convexity argument but I didn’t figure anything.
 A: It's equivalent to demonstrating :
$$2\prod_{m=0}^{k-1}(1-\frac{m}{n})\geq\frac{2n+k-k^2}{n}$$
By induction, you may prove the following inequality :
$$\prod_{i=1}^{n}(1-\alpha_{i})\geq 1-\sum_{i=1}^{n}\alpha_{i}$$
where $\alpha_{i}$ are positive. #in fact, we need $α_i∈[0,1]$. Thanks for the comment of "Thinking" below#
Applying this inequality, we obtain 
$$2\prod_{m=0}^{k-1}(1-\frac{m}{n})\geq 2(1-\sum_{m=0}^{k-1}\frac{m}{n}) = \frac{2n+k-k^2}{n}$$.
A: Here is a probabilistic way of proving the inequality.
As @曾靖國 noticed we have : 
$$2 \left ( 1 - \sum_{m=0}^{k-1} \frac{m}{n} \right) = \frac{2n+k-k^2}{n}$$
and thus the inequality we want to prove can  be rewritten as : 
$$2 \prod_{m = 0}^{k-1} \left ( 1- \frac{m}{n} \right ) \geq 2 \left ( 1 - \sum_{m=0}^{k-1} \frac{m}{n} \right) $$
Now let's consider indepent events : $A_0, ..., A_{k-1}$ and a probability $\mathbb{P}$ such that : 
$$\mathbb{P} \left ( A_i \right) = \frac{i}{n}$$
Hence we have (since the $\bar{A_i}$ are independent) : 
$$\mathbb{P} \left ( \bigcap_{i = 0}^{k-1} \bar{A_i} \right) = \prod_{m = 0}^{k-1} \left ( 1- \frac{m}{n} \right )$$
Thus : 
$$\mathbb{P} \left ( \overline{\bigcap_{i = 0}^{k-1} \bar{A_i}} \right) = \mathbb{P} \left ( \bigcup_{ i = 0}^{k-1} A_i \right) \leq \sum_{i = 0}^{k-1}  \mathbb{P} \left ( A_i \right ) = \sum_{m = 0}^{k-1} \frac{m}{n} $$
Since : 
$$\mathbb{P} \left ( \overline{\bigcap_{i = 0}^{k-1} \bar{A_i}} \right) = 1 - \mathbb{P} \left ( \bigcap_{i = 0}^{k-1} \bar{A_i} \right)$$
We have the desired inequality : 
$$2 \prod_{m = 0}^{k-1} \left ( 1- \frac{m}{n} \right ) \geq 2 \left ( 1 - \sum_{m=0}^{k-1} \frac{m}{n} \right) $$
Using the same probabilistic argument (or simply using induction as suggest by @曾靖國) , we can prove more generally that if : $\alpha_i \in [0,1]$ then : 
$$\prod_{i = 1}^n \left (1-\alpha_i \right ) \geq 1 - \sum_{i = 1}^n \alpha_i$$
A: $$
\begin{align}
\frac{n!}{(n-k)!\,n^k}
&=\prod_{j=0}^{k-1}\left(1-\frac jn\right)^{-1}\\
&\ge\prod_{j=0}^{k-1}\left(1+\frac jn\right)\\
&\ge1+\sum_{j=0}^{k-1}\frac jn\\
&=1+\frac{k^2-k}{2n}
\end{align}
$$
Thus, the inequality in the question should probably be the stronger
$$
2n! \geq (n-k)! n^{k-1} (2n-k+k^2)
$$
A: I may present another way to solve the problem, using Combinatorial argument and Induction.
The inequality can be written as:
$$ (n)(n-1)(n-2)…(n-(k-1)) \ge n^{k} - \frac{(k)(k-1)}{2} n^{k-1} $$
And let $f_{k} = (n)(n-1)(n-2)…(n-(k-1))$. Now if $n < \frac{(k)(k-1)}{2}$, then the right hand side is negative and proof is done. 
If $n \ge \frac{k(k-1)}{2}$: we will use induction on $k$ to prove it. 
Starting at $k=0,1,2$, it is clear the inequality is true. Now for $2 \le k = m \le n$, assume that the inequality is true
$$ f_{m} = (n)(n-1)(n-2)…(n-(m-1)) \ge n^{m} - \frac{(m)(m-1)}{2} n^{m-1} $$
and we wish to prove for $k=m+1 \le n$:
$$ (n)(n-1)(n-2)…(n-(m)) \ge n^{m+1} - \frac{(m+1)(m)}{2} n^{m} $$
Now, if wee see the terms as counting problem: we notice that $f_{m}$ is the same as the number of arrangements of $n$ different objects in $m$ positions. And notice that $n^{m}$ is the same thing but with repetition allowed. We also know that:
$$ f_{m} = n^{m} - \alpha_{m} $$
where $\alpha_{m}$ is the number of ways there is at least one repetition in the arrangement. So to prove the inequality is the same as proving 
$$ \alpha_{m+1} \le  \frac{(m+1)(m)}{2} n^{m}$$
knowing already that $ \alpha_{m} \le  \frac{(m)(m-1)}{2} n^{m-1}$. 
Now, since $\alpha_{m+1}$ is the number of arrangements where at least one repetition appears, we can use the result of $\alpha_{m}$ with one position left to be filled with $n$ choices,... plus $f_{m}$. So it must be $\alpha_{m+1} = n \alpha_{m} + f_{m} = (n-1)\alpha_{m} + n^{m} $. Thus
$$ \alpha_{m+1} \le (n-1)\frac{(m)(m-1)}{2} n^{m-1} + n^{m}  
= \frac{(m)(m-1)}{2} n^{m} + n^{m} - \frac{(m)(m-1)}{2} n^{m-1} =  \frac{(m^{2}-m + 2}{2} n^{m} - \frac{(m)(m-1)}{2} n^{m-1} \le \frac{(m^{2}-m + 2}{2} n^{m} \le
\frac{(m^{2}+m)}{2} n^{m} 
 $$
