Example 7 Determine the smallest natural number $k$, such that for any $a \in[0,1]$ and any $n \in \mathbf{N}$, we have $a^{k}(1-a)^{n}<\frac{1}{(n+1)^{3}}$.
Solve: Eliminate one parameter $a$, then find the minimum value of $k$. By the arithmetic-geometric mean inequality, we have
$$
\begin{array}{l}
\sqrt[n+k]{a^{k}\left[\frac{k}{n}(1-a)\right]^{n}} \leqslant \frac{k a+n\left[\frac{k}{n}(1-a)\right]}{n+k}=\frac{k}{n+k}. \\
\text { Thus, } a^{k}(1-a)^{n} \leqslant \frac{k^{k} n^{n}}{(n+k)^{n+k}},
\end{array}
$$

The equality holds if and only if $a=\frac{k(1-a)}{n}$ (i.e., $a=\frac{k}{n+k}$). This means that when $a=\frac{k}{n+k} (a \in [0,1])$, $a^{k}(1-a)^{n}$ attains its maximum value. Therefore, the problem can be transformed into: Determine the smallest natural number $k$ such that for any $n \in \mathbf{N}$,
$$
\frac{k^{k} n^{n}}{(n+k)^{n+k}} < \frac{1}{(n+1)^{3}}.
$$

It is easy to verify that when the pairs $(k, n)=(1,1),(2,1),(3,3)$, inequality (1) does not hold. Therefore, for it to hold for all $n \in \mathbf{N}$, we must have $k \geqslant 4$.
Next, we prove that when $k=4$, inequality (*) holds, i.e., for all $n \in \mathbf{N}$, we have $4^{4} n^{n}(n+1)^{3} < (n+4)^{n+4}$.
In fact, when $n=1,2,3$, it can be directly verified.
When $n \geqslant 4$, we have

i.e., $4^{4} n^{n}(n+1)^{3} < (n+4)^{n+4}$.
In summary, the minimum value of $k$ is 4.