In the expansion of $(1+x)^{15}$ the coefficient of
$$
x^{10}={ }^{15} C_{10} \text {. }
$$

And the expansion of
$$
(1-x)^{-n}=1+n x+\frac{n(n+1)}{2 !} x^2+\frac{n(n+1)(n+2) x^3}{3 !}+\ldots
$$

So, coefficient of $x^5=\frac{n(n+1)(n+2)(n+3)(n+4)}{5 !}$
According to the question,
$$
\frac{n(n+1)(n+2)(n+3)(n+4)}{5 !}={ }^{15} C_{10}
$$


$$
\begin{gathered}
=\frac{15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \\
\Rightarrow \frac{n(n+1)(n+2)(n+3)(n+4)}{5 !}=7 \times 13 \times 3 \times 11 \\
\Rightarrow n(n+1)(n+2)(n+3)(n+4)=5 !(7 \times 13 \times 3 \times 11) \\
\Rightarrow n(n+1)(n+2)(n+3)(n+4)=11 \times 12 \times 13 \times 14 \times 15
\end{gathered}
$$

By comparing, we get
$$
n=11
$$