4.5.15 Find the smallest positive integer $k$, such that for all $a$ satisfying $0 \leqslant a \leqslant 1$ and all positive integers $n$, the inequality holds: $a^{k}(1-a)^{n}<\frac{1}{(n+1)^{3}}$.
We first try to eliminate one parameter $a$, and then find the minimum value of $k$. Using the mean inequality, we get
$$
\sqrt[n+k]{a^{k}\left[\frac{k}{n}(1-a)\right]^{n}} \leqslant \frac{k a+n\left[\frac{k}{n}(1-a)\right]}{k+n}=\frac{k}{k+n},
$$

Therefore, $a^{k}(1-a)^{n} \leqslant \frac{k^{k} n^{n}}{(n+k)^{n+k}}$, with equality holding 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 $n=1$, (1) becomes $\frac{1}{2^{2}}<\frac{1}{2^{3}}$, which is a contradiction. When $k=2$, taking $n=1$, (1) becomes $\frac{4}{27}<\frac{1}{8}$, which is a contradiction. When $k=3$, taking $n=3$, (1) becomes $\frac{1}{64}<\frac{1}{64}$, which is a contradiction. Therefore, $k \geqslant 4$. We now prove that when $k=4$, (1) 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 (2) holds. When $n \geqslant 4$, using the mean inequality, we get
$$
\begin{aligned}
\sqrt[n+1]{4^{4} n^{n}(n+1)^{3}} & =\sqrt[n+4]{16(2 n)(2 n)(2 n)(2 n) n^{n-4}(n+1)^{3}} \\
& \leqslant \frac{16+4 \cdot 2 n+n(n-4)+3(n+1)}{n+4} \\
& =\frac{n^{2}+7 n+19}{n+4}<\frac{n^{2}+8 n+16}{n+4}=n+4.
\end{aligned}
$$

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