Analysis: The left algebraic expression is similar to the expansion of the binomial theorem, which can be associated with the binomial theorem, and the problem can be solved easily.
Proof: $\because \mathrm{C}_{n}^{1}+\mathrm{C}_{n}^{2}+\cdots+\mathrm{C}_{n}^{n}$
$$\begin{aligned} & =2^{n}-1=\frac{2^{n}-1}{2-1} \\ & =1+2+2^{2}+2^{3}+\cdots+2^{n-1} \\ & >n \sqrt[n]{2^{1+2+\cdots+(n-1)}} \\ & =n \cdot 2^{\frac{n-1}{2}}, \\ \therefore \mathrm{C}_{n}^{1}+ & \mathrm{C}_{n}^{2}+\cdots+C_{n}^{n}>n \cdot 2^{\frac{n-1}{2}}(n \geqslant 2) . \end{aligned}$$