Five, (20 points) Find the smallest natural number $k$, such that for any $x \in [0,1]$ and $n \in \mathbf{N}_{+}$, the inequality
$$
x^{k}(1-x)^{n}<\frac{1}{(1+n)^{3}}
$$

always holds.
Given $x \in [0,1]$, by the AM-GM inequality, we have
$$
\begin{array}{l}
x^{k}(1-x)^{n} \\
=\underbrace{x \cdots \cdots x}_{k \uparrow} x \cdot \underbrace{(1-x) \cdot(1-x) \cdots \cdots(1-x)}_{n \uparrow} \\
=\left(\frac{n}{k}\right)^{n} \underbrace{x \cdots \cdots x}_{k \uparrow} \cdot \underbrace{\frac{k(1-x)}{n} \cdots \cdots \cdot \frac{k(1-x)}{n}}_{n \uparrow} \\
\leqslant \frac{n^{n}}{k^{n}}\left[\frac{k x+k(1-x)}{n+k}\right]^{n+k} \\
=\frac{n^{n} k^{k}}{(n+k)^{n+k}} .
\end{array}
$$

When $x=\frac{k}{n+k}$, $x^{k}(1-x)^{n}$ reaches its maximum value of
$$
\frac{n^{n} k^{k}}{(n+k)^{n+k}}.
$$

Thus, the problem is equivalent to finding the smallest natural number $k$ such that
$$
\frac{(k+n)^{k+n}}{k^{k} n^{n}}>(n+1)^{3} \quad\left(n \in \mathbf{N}_{+}\right)
$$

holds for all $n \in \mathbf{N}_{+}$. Testing $(k, n)=(1,1),(2,1),(3,3)$, we find that $k=1,2,3$ do not satisfy the condition.
When $k=4$, if $n=1$, inequality (1) clearly holds.
If $n \geqslant 2$, then
$$
\begin{array}{l}
4^{4} n^{n}(n+1)^{3}=n^{n-2}(2 n)^{2}(2 n+2)^{3} \times 2^{3} \\
\leqslant\left[\frac{(n-2) n+2 \times 2 n+3(2 n+2)+2^{3}}{n+4}\right]^{n+4} \\
=\left(\frac{n^{2}+8 n+14}{n+4}\right)^{n+4}<\left(\frac{n^{2}+8 n+16}{n+4}\right)^{n+4} \\
=(n+4)^{n+4} .
\end{array}
$$

Therefore, inequality (1) holds. Hence, $k=4$ is the solution.