Elegant proof that $\left|(1-x)(2-x)(3-x)\cdots(n-x)\right| < n!$ for all $0I am looking for a nice argument which shows that
$$\left|\,(1-x)(2-x)(3-x)\cdots(n-x)\,\right| < n!$$
for all $0 < x < n+1$.
I've proven it already (cases based on the integer part of $x$), but am unsatisfied with my proof.
My proof, by request:
If $x \in \mathbb{Z}$ then the LHS is zero.
Therefore, let $k < x < k+1$ with $0 \leq k \leq n$ and $k \in \mathbb{Z}$.
Then
$$\begin{align}
&|(1-x)(2-x)(3-x)\cdots(n-x)| \\
&< |1-(k+1)| |2-(k+1)| \cdots |k-(k+1)|\,|(k+1)-k||(k+2)-k| \cdots |n-k| \\
&= |k||k-1| \cdots |1| \, |1| |2| \cdots |n-k| \\
&= |1| |2| \cdots |n-k| \, |1| \cdots |k-1| |k| \\
&\leq |1| |2| \cdots |n-k| |n-k+1| \cdots |n| \\
&= n!
\end{align}$$
I suspect there's a much nicer way to approach the problem, but I can't seem to find one.
Any help is much appreciated!
 A: Suppose
$$
\mid (1-x) (2-x) \cdots (n-x) \mid < n!
$$
for all $0 < x< n+1$.
If $0< x < n+1$, then
$$
\mid (1-x) (2-x) \cdots (n-x) (n+1-x) \mid < n! \mid n+1-x \mid < (n+1)!
$$
If $n+1<x<n+2$, then let $y=x-1$ so $n<y<n+1$
$$
\mid (1-x) (2-x) \cdots (n-x) (n+1-x) \mid = \mid (-y) (1-y) (2-y) \cdots (n-y) \mid \\
= \mid y \mid \mid (1-y) (2-y) \cdots (n-y) \mid\\
< \mid y \mid n!\\
< (n+1)!
$$
That was the induction step, let's go back to the base case.
$$
0 < x < 0+1\\
0 < 1-x < 1\\
\mid (1-x) \mid = (1-x) < 1!=1
$$
A: You can use induction here, due to the symmetry of the expression.
Suppose that $$\left|(1-x)(2-x)(3-x)\cdots(n-1-x)\right| < (n-1)!$$ for $0 < x < n$.
Then if $0 < x < n$, we have $|n-x|< n$, so $$|(1-x)(2-x)(3-x)\cdots(n-x)| < n!$$ by the inductive hypothesis.
And if $n\le x<n+1$, put $y=n+1-x$; now $0<y\le 1$, so we can use the inductive hypothesis on $y$, to get $$|(1-y)(2-y)(3-y)\cdots(n-y)| < n!$$
But $|(1-y)(2-y)(3-y)\cdots(n-y)|$ is just $|(1-x)(2-x)(3-x)\cdots(n-x)|$, with the terms reversed, so we are done.
A: What you have is a nice, direct  proof. I would only get rid of the
absolute values, so that it becomes more obvious how each
term is estimated.
So for $x \in (k, k+1)$ with an integer $k$, $0 \le k \le n$, we have
$$
|(1-x)(2-x)(3-x)\cdots(n-x)| \\
 = (x-1)(x-2)\cdots(x-k) \, \times \, (k+1-x)(k+2-x) \cdots (n-x) \\
< (k)(k-1)\cdots (1) \, \times \, (1)(2) \cdots(n-k) \, .
$$
In the second part of the product we can increase each factor by $k$, 
so that the expression is
$$
 \le (k)(k-1)\cdots (1) \, \times \, (k+1)(k+2) \cdots(n) = n! \, .
$$
Alternatively use that
$$
(k)(k-1)\cdots (1) \, \times \, (1)(2) \cdots(n-k) = k!(n-k)! = \frac{n!}{\binom nk} \le n! \, .
$$
