Let $n$ ($n\ge 3$) be a natural number. Denote by $f(n)$ the least natural number by which $n$ is not divisible (e.g. $f(12)=5$). If $f(n)\ge 3$, we may have $f(f(n))$ in the same way. Similarly, if $f(f(n))\ge 3$, we may have $f(f(f(n)))$, and so on. If $\underbrace{f(f(\dots f}_{k\text{ times}}(n)\dots ))=2$, we call $k$ the “[i]length[/i]” of $n$ (also we denote by $l_n$ the “[i]length[/i]” of $n$). For arbitrary natural number $n$ ($n\ge 3$), find $l_n$ with proof.
To find the "length" \( l_n \) of a natural number \( n \ge 3 \), we need to analyze the function \( f(n) \), which is defined as the least natural number by which \( n \) is not divisible. We will consider two cases: when \( n \) is odd and when \( n \) is even.

1. **Case 1: \( n \) is odd**

   If \( n \) is odd, the smallest natural number by which \( n \) is not divisible is \( 2 \). Therefore, \( f(n) = 2 \). Since \( f(n) = 2 \), the length \( l_n \) is \( 1 \).

2. **Case 2: \( n \) is even**

   If \( n \) is even, we need to find the smallest natural number by which \( n \) is not divisible. Let's denote \( n = 2^a \cdot b \), where \( b \) is an odd number and \( a \ge 1 \).

   - If \( n = 2^m \) for some \( m \ge 1 \), then \( f(n) = 3 \) because \( 2^m \) is not divisible by \( 3 \). Next, we need to find \( f(f(n)) = f(3) \). Since \( 3 \) is odd, \( f(3) = 2 \). Therefore, the length \( l_n \) is \( 2 \).

   - If \( n \) is not a power of \( 2 \), then \( n \) has an odd factor \( b \). The smallest natural number by which \( n \) is not divisible is \( 3 \) (since \( n \) is even and not divisible by \( 3 \)). Therefore, \( f(n) = 3 \). Next, we need to find \( f(f(n)) = f(3) \). Since \( 3 \) is odd, \( f(3) = 2 \). Therefore, the length \( l_n \) is \( 2 \).

In summary, the length \( l_n \) of a natural number \( n \ge 3 \) is determined as follows:
- If \( n \) is odd, \( l_n = 1 \).
- If \( n \) is even, \( l_n = 2 \).

The final answer is \( \boxed{ l_n = 1 } \) if \( n \) is odd, and \( l_n = 2 \) if \( n \) is even.