[color=darkred] Let $m$ and $n$ be two nonzero natural numbers. Determine the minimum number of distinct complex roots of the polynomial $\prod_{k=1}^m\, (f+k)$ , when $f$ covers the set of $n^{\text{th}}$ - degree polynomials with complex coefficients.
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1. **Understanding the Problem:**
   We need to determine the minimum number of distinct complex roots of the polynomial \(\prod_{k=1}^m (f+k)\), where \(f\) is an \(n^{\text{th}}\)-degree polynomial with complex coefficients.

2. **Analyzing the Polynomials:**
   Consider the polynomials \(f+k\) for \(k = 1, 2, \ldots, m\). Each \(f+k\) is an \(n^{\text{th}}\)-degree polynomial. Therefore, each \(f+k\) has exactly \(n\) roots (counting multiplicities).

3. **Distinct Roots:**
   Observe that no two polynomials \(f+i\) and \(f+j\) (for \(i \neq j\)) can have a common root. This is because if \(r\) were a common root, then \(f(r) + i = 0\) and \(f(r) + j = 0\), which implies \(i = j\), a contradiction.

4. **Counting Roots:**
   Since each \(f+k\) has \(n\) roots and there are \(m\) such polynomials, the total number of roots (counting multiplicities) is \(nm\).

5. **Repeated Roots:**
   If there are repeated roots, they must be repeated within a single polynomial \(f+k\). Let \(x_1, x_2, \ldots, x_t\) be the repeated roots of \(\prod_{k=1}^m (f+k)\), where \(x_j\) has multiplicity \(a_j\).

6. **Derivative Condition:**
   The \(k^{\text{th}}\) order root of \(f+k\) is the \((k-1)^{\text{th}}\) order root of \(f'\). Since \(\deg(f') = n-1\), we have:
   \[
   (a_1 - 1) + (a_2 - 1) + \cdots + (a_t - 1) \leq n - 1
   \]
   Simplifying, we get:
   \[
   a_1 + a_2 + \cdots + a_t \leq n - 1 + t
   \]

7. **Minimum Number of Distinct Roots:**
   The number of distinct roots of \(\prod_{k=1}^m (f+k)\) is:
   \[
   nm - (a_1 + a_2 + \cdots + a_t) + t
   \]
   Using the inequality \(a_1 + a_2 + \cdots + a_t \leq n - 1 + t\), we get:
   \[
   nm - (n - 1 + t) + t = nm - n + 1
   \]
   Therefore, the minimum number of distinct roots is:
   \[
   n(m-1) + 1
   \]

8. **Equality Case:**
   Equality occurs when \(f(x) = (x-1)^n - 1\), which ensures that the roots are distributed in such a way that the minimum number of distinct roots is achieved.

The final answer is \(\boxed{n(m-1) + 1}\)