Let $ a > b > 1, b$ is an odd number, let $ n$ be a positive integer. If $ b^n|a^n\minus{}1,$ then $ a^b > \frac {3^n}{n}.$
1. Given \( a > b > 1 \) and \( b \) is an odd number, let \( n \) be a positive integer. We are given that \( b^n \mid a^n - 1 \). We need to show that \( a^b > \frac{3^n}{n} \).

2. Without loss of generality, assume \( b \) is a prime number. This is because if \( b \) is not a prime, we can consider its prime factors, and the argument will still hold.

3. By the properties of \( v_p \) (the p-adic valuation), we have \( v_p(a^n - 1) \geq n \). This is because \( b^n \mid a^n - 1 \) implies that \( v_p(a^n - 1) \geq n \).

4. Let \( d \) be the order of \( a \) modulo \( p \). By the Lifting The Exponent (LTE) lemma, we have:
   \[
   v_p(a^n - 1) = v_p(a^d - 1) + v_p\left(\frac{n}{d}\right)
   \]
   Since \( d \mid p-1 \), we know that \( d \) is a divisor of \( p-1 \).

5. From the LTE lemma, we have:
   \[
   n \leq v_p(a^n - 1) = v_p(a^d - 1) + v_p(n)
   \]
   This implies:
   \[
   p^{n - v_p(n)} \leq a^d - 1
   \]

6. Since \( d \mid p-1 \), we have \( a^d - 1 < a^p \). Therefore:
   \[
   p^{n - v_p(n)} \leq a^d - 1 < a^p
   \]

7. Thus, we get:
   \[
   a^p > \frac{p^n}{p^{v_p(n)}} = \frac{p^n}{n}
   \]

8. Since \( b \) is an odd prime and \( b \geq 3 \), we have:
   \[
   a^b > \frac{b^n}{n} \geq \frac{3^n}{n}
   \]

Therefore, we have shown that \( a^b > \frac{3^n}{n} \).

The final answer is \( \boxed{ a^b > \frac{3^n}{n} } \).