Let consider the following function set $$F=\{f\ |\ f:\{1,\ 2,\ \cdots,\ n\}\to \{1,\ 2,\ \cdots,\ n\} \}$$
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[*] Find $|F|$
[*] For $n=2k$ prove that $|F|< e{(4k)}^{k}$
[*] Find $n$, if $|F|=540$ and $n=2k$
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1. **Finding $|F|$:**

   The set $F$ consists of all functions from the set $\{1, 2, \cdots, n\}$ to itself. The number of such functions is given by $n^n$ because each of the $n$ elements in the domain can be mapped to any of the $n$ elements in the codomain.

   Therefore, 
   \[
   |F| = n^n
   \]

2. **Proving $|F| < e(4k)^k$ for $n = 2k$:**

   Given $n = 2k$, we need to show that $n^{n} < e(4k)^k$.

   Substituting $n = 2k$ into $n^n$, we get:
   \[
   (2k)^{2k}
   \]

   We need to show:
   \[
   (2k)^{2k} < e(4k)^k
   \]

   Simplifying the right-hand side:
   \[
   e(4k)^k = e \cdot (4k)^k
   \]

   Now, we compare $(2k)^{2k}$ and $e \cdot (4k)^k$:
   \[
   (2k)^{2k} = (2k)^{2k} = (2^2 \cdot k^2)^k = 4^k \cdot k^{2k}
   \]

   We need to show:
   \[
   4^k \cdot k^{2k} < e \cdot 4^k \cdot k^k
   \]

   Dividing both sides by $4^k$:
   \[
   k^{2k} < e \cdot k^k
   \]

   Dividing both sides by $k^k$:
   \[
   k^k < e
   \]

   This inequality holds for $k > 2$. Therefore, we have shown that:
   \[
   (2k)^{2k} < e(4k)^k
   \]

3. **Finding $n$ if $|F| = 540$ and $n = 2k$:**

   Given $|F| = 540$, we know:
   \[
   n^n = 540
   \]

   Since $n = 2k$, we substitute $n$:
   \[
   (2k)^{2k} = 540
   \]

   We need to find $k$ such that this equation holds. Let's test some values of $k$:

   - For $k = 1$:
     \[
     (2 \cdot 1)^{2 \cdot 1} = 2^2 = 4 \quad (\text{not } 540)
     \]

   - For $k = 2$:
     \[
     (2 \cdot 2)^{2 \cdot 2} = 4^4 = 256 \quad (\text{not } 540)
     \]

   - For $k = 3$:
     \[
     (2 \cdot 3)^{2 \cdot 3} = 6^6 = 46656 \quad (\text{not } 540)
     \]

   Since none of these values satisfy the equation, there is no integer $k$ such that $(2k)^{2k} = 540$. Therefore, there is no such $n$ that satisfies the given condition.

The final answer is $\boxed{None}$