For any natural number $n > 1$ write the finite decimal expansion of $\frac{1}{n}$ (for example we write $\frac{1}{2}=0.4\overline{9}$ as its infinite decimal expansion not $0.5)$. Determine the length of non-periodic part of the (infinite) decimal expansion of $\frac{1}{n}$.
1. **Understanding the Problem:**
   We need to determine the length of the non-periodic part of the decimal expansion of \(\frac{1}{n}\) for any natural number \(n > 1\). 

2. **Prime Factorization and Decimal Expansion:**
   Let \(v_p(n)\) be the largest power of a prime \(p\) that divides \(n\). The answer is \(\max[v_2(n), v_5(n)]\).

3. **Case 1: \(\gcd(n, 10) = 1\):**
   For \(n\) such that \(\gcd(n, 10) = 1\), we need to show that \(\frac{1}{n}\) has no non-repeating part. 

   When dividing 1 by \(n\), the integer part of the quotient is 0. After placing the decimal point, the new dividend becomes 10. The first digit of the fractional part of the quotient is \(n_1\), where \(n_1\) is the largest multiple of \(n\) less than 10. The process continues similarly for subsequent digits.

4. **Sequence of Steps:**
   We can write this process as a sequence of steps:
   \[
   (10, n_1n) \mapsto (10^2 - 10n_1n, n_2n) \mapsto (10^3 - 10^2n_1n - 10n_2n, n_3n) \mapsto \cdots
   \]
   In each step, we choose \(n_i\) such that the first number of the pair is as small as possible for a fixed \(n\).

5. **Lemma:**
   There exists an \(x\) such that the remainder after the \(x\)-th step is 1.

6. **Proof of Lemma:**
   Choose \(x\) such that the smallest repunit (numbers of the form \(111\ldots1\)) with \(x\) digits is divisible by \(n\). Call this repunit \(r_x\). We can always do so as long as \(\gcd(n, 10) = 1\).

   After \(x\) steps, the \(x\)-th pair is:
   \[
   (10^x - n(10^{x-1}n_1 + 10^{x-2}n_2 + \cdots + 10n_{x-1}), n_xn)
   \]
   The remainder after the \(x\)-th step would be:
   \[
   10^x - n(10^{x-1}n_1 + 10^{x-2}n_2 + \cdots + 10n_{x-1} + n_x)
   \]
   For this to be 1, we need:
   \[
   n(10^{x-1}n_1 + 10^{x-2}n_2 + \cdots + 10n_{x-1} + n_x) = 10^x - 1
   \]
   which is \(9r_x\).

7. **Verification:**
   Let \(n(10^{x-1}n_1 + 10^{x-2}n_2 + \cdots + 10n_{x-1} + n_x) = s\). We wish to show that it is possible to have \(\frac{s}{n} = 9\frac{r_x}{n}\). 

   Since \(10^x - s\) is positive and as small as possible, \(s\) is as large as possible. We also have \(s < 10^x\), meaning \(s\) must have \(x\) digits. The largest number with \(x\) digits is \(9r_x\). 

   Since \(r_x < 10^x\), \(\frac{r_x}{n} < 10^{x-1}\), and \(9\frac{r_x}{n} < 10 \cdot 10^{x-1} = 10^x\). Thus, the number of digits of \(9\frac{r_x}{n}\) is less than or equal to \(x\). 

   Therefore, \(s = n(10^{x-1}n_1 + 10^{x-2}n_2 + \cdots + 10n_{x-1}) = 9r_x = 10^x - 1\). So the remainder after the \(x\)-th step is \(10^x - (10^x - 1) = 1\).

8. **Conclusion for \(\gcd(n, 10) = 1\):**
   The digits of the quotient start repeating from \(n_1\), and the length of the non-repeating part is 0.

9. **Case 2: \(n = 2^a 5^b k\) where \(a, b \geq 0\) and \(\gcd(k, 10) = 1\):**
   \[
   \frac{1}{n} = \frac{1}{2^a 5^b} \cdot \frac{1}{k}
   \]
   Since the non-repeating part of \(\frac{1}{k}\) is 0, the non-repeating part of \(\frac{1}{n}\) is simply \(\max(a, b)\).

Thus, the length of the non-periodic part of the decimal expansion of \(\frac{1}{n}\) is \(\max[v_2(n), v_5(n)]\).

The final answer is \( \boxed{ \max[v_2(n), v_5(n)] } \).