$$\begin{array}{l} \text {Prove } f(n-1)-1 \\ =f(n)(f(n)-1) . \end{array}$$

Since $f(1)=2$, from the above equation, we know that $f(n)$ is monotonically increasing.
$$\begin{array}{l} \frac{1}{f(n+1)-1}=\frac{1}{f(n)(f(n)-1)} \\ =\frac{1}{f(n)-1}-\frac{1}{f(n)} . \end{array}$$

Thus, we have
$$\frac{1}{f(n)}=\frac{1}{f(n)-1}-\frac{1}{f(n+1)-1} .$$

Using the above equation, we get
$$\sum_{k=1}^{n} \frac{1}{f(k)}=1-\frac{1}{f(n+1)-1} .$$

We will prove the following by induction:
2^{2^{n-1}}2^{2^{k-1}} 2^{2^{2-1}}=2^{2^{k}} . \end{array}

Since $f(k)$ is a positive integer $(k \in N)$,
$$\begin{array}{l} f(k+1)-1 \leqslant 2^{2^{k}}-1 . \\ f(k+2)-1=f(k+1)(f(k+1)-1) \\ \leqslant 2^{2 k}\left(2^{2 k}-1\right)<2^{2 k+1} . \end{array}$$
(*) holds for any positive integer $n$. Therefore, the desired inequality is immediately obtained.