Prove that representing $n$ in binary as
$$\begin{array}{l} n=2^{\alpha_{1}}+2^{\alpha_{2}}+\cdots+2^{\alpha_{k}}\left(\alpha_{1} \geqslant \alpha_{2} \geqslant \cdots \geqslant \alpha_{k}\right) \\ \Rightarrow(1+x)^{n}=(1+x)^{2 \alpha_{1}}(1+x)^{2 \alpha_{2}} \cdots(1+x)^{2 \alpha_{k}} \\ \Rightarrow(1+x)^{n} \\ \quad \equiv\left(1+x^{2 \alpha_{1}}\right)\left(1+x^{2 a_{2}}\right) \cdots\left(1+x^{2 \alpha_{k}}\right)(\bmod 2) . \end{array}$$

The polynomial on the right side of the equation contains exactly $2^{k}$ terms, all of whose coefficients are 1 (odd), meaning that in the expansion of $(1+x)^{n}$, exactly $2^{k}$ terms have coefficients $\mathrm{C}_{n}^{m}$ that are odd.