How to get a bound of this function?

let $$f(x)=\prod_{i=0}^n\left(x-\frac{i}{n}\right)$$

where $x\in[0,1]$.

How to deduce a bound for it? Particularly, I want to prove $$|f(x)|\leq\frac{(n+1)!}{n^{n+1}}$$

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We can do a bit better. Suppose we have $n+1$ points $x_0, \cdots, x_n$ distributed evenly along $[a,b]$ with $x_0=a,x_n=b.$ Then for $x\in [a,b]$, $$\left|\prod_{i=0}^n (x-x_i)\right|\leq\frac{n!}{4}\left(\frac{b-a}{n}\right)^{n+1} .$$
For your problem this gives $|f(x)|\leq \dfrac{n!}{4n^{n+1}}.$ The bound in the OP is $(n+1)/4$ times larger than this bound.
Hints: By scaling and shifting, it is equivalent to show that for $x\in [0,n]$ we have $|P(x)|\leq n!/4$ where $P(x)=x(x-1)\cdots (x-n).$ To show this, use induction. Note, in the induction step the inequality of the induction hypothesis does not apply over the entire interval for the next case, but we get around this by exploiting some symmetry of $P(x).$
Hint: start by writing each term as $\frac{nx-i}n$