Example 2 Determine the smallest natural number $k$, such that for any $a \in [0,1]$ and any $n \in \mathbf{N}$ we have
$$
a^{k}(1-a)^{n}<\frac{1}{(n+1)^{3}} .
$$
Solution: By the arithmetic-geometric mean inequality, we have
$$
\begin{array}{l}
\sqrt[n+k]{a^{k}\left[\frac{k}{n}(1-a)\right]^{n}} \\
\leqslant \frac{k a+n\left[\frac{k}{n}(1-a)\right]}{n+k}=\frac{k}{n+k} .
\end{array}
$$

Thus, $a^{k}(1-a)^{n} \leqslant \frac{k^{k} n^{n}}{(n+k)^{n+k}}$.
Equality holds if and only if $a=\frac{k(1-a)}{n}\left(\right.$ i.e., $\left.a=\frac{k}{n+k}\right)$.

This means that when $a=\frac{k}{n+k}(\in[0,1])$, $a^{k}(1-a)^{n}$ attains its maximum value. Therefore, the problem can be reformulated as: 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 for 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 (1) holds, i.e., for all $n \in \mathbf{N}$, we have $4^{4} n^{n}(n+1)^{3}<(n+4)^{n+4}$.
Indeed, for $n=1,2,3$, this can be directly verified.
For $n \geqslant 4$, we have
$$
\begin{array}{l}
\sqrt[n+1]{4^{4} n^{n}(n+1)^{3}} \\
=\sqrt[n+4]{16 \times(2 n)(2 n)(2 n)(2 n) n^{n-4}(n+1)^{3}} \\
\leqslant \frac{16+8 n+(n-4) n+3(n+1)}{n+4} \\
=\frac{n^{2}+7 n+19}{n+4}<\frac{n^{2}+8 n+16}{n+4}=n+4,
\end{array}
$$

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