Example 6 Find the smallest positive integer $k$, such that for all $a$ satisfying $0 \leqslant a \leqslant 1$ and all positive integers $n$, we have
$$
a^{k}(1-a)^{n}<\frac{1}{(n+1)^{3}} .
$$
Solution: First, we aim to eliminate the parameter $a$, and then it will be easier to find the minimum value of $k$. Using the arithmetic-geometric mean inequality, we get
$$
\begin{array}{l}
\sqrt[n+k]{a^{k}\left[\frac{k}{n}(1-a)\right]^{n}} \\
\leqslant \frac{k a+n\left[\frac{k}{n}(1-a)\right]}{k+n}=\frac{k}{k+n} .
\end{array}
$$

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

Thus, we need to find the smallest positive integer $k$ such that for any positive integer $n$, we have
$$
\frac{k^{k} n^{n}}{(n+k)^{n+k}}<\frac{1}{(n+1)^{3}}.
$$

When $k=1$, taking $n=1$ leads to a contradiction in equation (4);
When $k=2$, taking $n=1$ leads to a contradiction in equation (4);
When $k=3$, taking $n=3$ leads to a contradiction in equation (4).
Therefore, $k \geqslant 4$.
Next, we prove that when $k=4$, equation (4) holds, i.e.,
$$
4^{4} n^{n}(n+1)^{3}<(n+4)^{n+4}.
$$

When $n=1,2,3$, it is easy to verify that equation (5) holds.
When $n \geqslant 4$, using the arithmetic-geometric mean inequality, we get
$$
\begin{array}{l}
\sqrt[n+4]{4^{4} n^{n}(n+1)^{3}} \\
=\sqrt[n+4]{16(2 n)(2 n)(2 n)(2 n) n^{n-4}(n+1)^{3}} \\
\leqslant \frac{16+4 \times 2 n+n(n-4)+3(n+1)}{n+4} \\
=\frac{n^{2}+7 n+19}{n+4}<n+4 .
\end{array}
$$

In conclusion, the minimum value of $k$ is 4.