Combinatorial proof of ${2n\choose n}n!\le (2n)^{n}$

The lefthand side of the inequality in the title is the number of ways to put $n$ balls numbered $1$ through $n$ into $2n$ slots numbered $1$ through $2n$ if no slot is allowed to contain more than one ball. The righthand side is the number of ways to put the same $n$ balls into the same $2n$ slots with no restrictions on how many balls can occupy any slot.

Consider the ratio:

$$\frac{2n \times 2n \times 2n \times \cdots \times 2n} {(2n) \times (2n-1) \times (2n-2) \times \cdots \times (2n + 1 - n)}.$$

Clearly, the numerator is bigger than the denominator.

Let $X = [n] = \{1,2,\ldots,n\}$, and $Y = [2n] = \{1,2,\ldots ,2n\}$.

Observe that $(2n)^n$ is the number of functions $f:X\to Y$. Indeed, for each $i\in X$, you have $2n$ choices to assign $f(i)$.

Next, note ${2n \choose n}n!$ is the number of injective functions $f:X\to Y$. Indeed, for an injective (one-to-one) function, we first pick an image for $f$, which is a size $n$ subset of $Y$. There are $2n\choose n$ many ways. Now with the image chosen, we have $n!$ ways to assign them to $f(i)$.

Since the set of all injective functions is a subset of all functions $X\to Y$, we have the desired inequality.

Note: In general ${|Y| \choose |X|} |X|!$ is the number of injective functions $X\to Y$ (whenever sensible), while $|Y|^{|X|}$ is the total number of functions $X\to Y$.