Example 2 Let the function $f: \mathbf{N}^{\top} \rightarrow \mathbf{N}^{+}, f(1)=1$, try to solve the functional inequality $f(n+1) \geqslant f(n)+2^{n}\left(n \in \mathbf{N}^{+}\right)$.
Solution 1 Introduce a non-negative function $p(n)\left(p(n) \geqslant 0, n \in \mathbf{N}^{+}\right)$, and let
$$
\begin{array}{l} 
f(n+1)=f(n)+2^{n}+p(n), \\
f(n)=[f(n)-f(n-1)]+[f(n-1)-f(n-2)]+\cdots+[f(2)-f(1)]+f(1) \\
= {\left[2^{n}+p(n-1)\right]+\left[2^{n-2}+p(n-2)\right]+\cdots+[1+p(1)]+1, }
\end{array}
$$

Therefore, the solution to the original functional inequality is $f(n)=2^{n}+\sum_{i=1}^{n} p(i)$, where $\forall p(n) \geqslant 0, n \in \mathbf{N}^{+}$.
Solution 2 The inequality $f(n+1) \geqslant f(n)+2^{n}$ can be transformed into $f(n+1)-2^{n+1} \geqslant f(n)-2^{n}$. Let $g(n)=f(n)-2^{n}$. Then we have $g(n+1) \geqslant g(n)$. Therefore, the solution to the given inequality is $f(n)=2^{n}+g(n)$, where $g(n+1) \geqslant g(n), n \in \mathbf{N}^{+}$.