 Let $p$ be a prime number, $n$ a natural number which is not divisible by $p$, and $\mathbb{K}$ is a finite field, with $char(K) = p, |K| = p^n, 1_{\mathbb{K}}$ unity element and $\widehat{0} = 0_{\mathbb{K}}.$ For every $m \in \mathbb{N}^{*}$ we note 
    $ \widehat{m} = \underbrace{1_{\mathbb{K}} + 1_{\mathbb{K}} + \ldots + 1_{\mathbb{K}}}_{m \text{ times}} $ and define the polynomial 

    \[
    f_m = \sum_{k = 0}^{m} (-1)^{m - k} \widehat{\binom{m}{k}} X^{p^k} \in \mathbb{K}[X].
    \]

    a) Show that roots of $f_1$ are $ \left\{ \widehat{k} | k \in \{0,1,2, \ldots , p - 1 \}  \right\}$.

    b) Let $m \in \mathbb{N}^{*}.$ Determine the set of roots from $\mathbb{K}$ of polynomial $f_{m}.$  
### Part (a)

To show that the roots of \( f_1 \) are \( \left\{ \widehat{k} \mid k \in \{0,1,2, \ldots , p - 1 \} \right\} \):

1. **Define \( f_1 \):**
   \[
   f_1(X) = \sum_{k=0}^{1} (-1)^{1-k} \widehat{\binom{1}{k}} X^{p^k}
   \]
   Simplifying the binomial coefficients:
   \[
   f_1(X) = (-1)^1 \widehat{\binom{1}{0}} X^{p^0} + (-1)^0 \widehat{\binom{1}{1}} X^{p^1}
   \]
   Since \(\widehat{\binom{1}{0}} = 1_{\mathbb{K}}\) and \(\widehat{\binom{1}{1}} = 1_{\mathbb{K}}\):
   \[
   f_1(X) = -1_{\mathbb{K}} X + 1_{\mathbb{K}} X^p
   \]
   Therefore:
   \[
   f_1(X) = X^p - X
   \]

2. **Find the roots of \( f_1(X) \):**
   We need to solve:
   \[
   X^p - X = 0
   \]
   Factoring out \(X\):
   \[
   X(X^{p-1} - 1) = 0
   \]
   The solutions are:
   \[
   X = 0 \quad \text{or} \quad X^{p-1} = 1
   \]
   In the finite field \(\mathbb{K}\) with \(|\mathbb{K}| = p^n\), the elements \(\widehat{k}\) for \(k \in \{0, 1, 2, \ldots, p-1\}\) satisfy \(X^{p-1} = 1\).

3. **Conclusion:**
   The roots of \( f_1(X) \) are:
   \[
   \left\{ \widehat{k} \mid k \in \{0, 1, 2, \ldots, p-1\} \right\}
   \]

### Part (b)

To determine the set of roots from \(\mathbb{K}\) of polynomial \( f_m \):

1. **Inductive Hypothesis:**
   Assume that the roots of \( f_{m-1}(X) \) are \(\mathbb{F}_p\).

2. **Expression for \( f_m(X) \):**
   \[
   f_m(X) = \sum_{k=0}^{m} (-1)^{m-k} \widehat{\binom{m}{k}} X^{p^k}
   \]

3. **Inductive Step:**
   We need to show that the roots of \( f_m(X) \) are also in \(\mathbb{F}_p\).

4. **Base Case:**
   For \( m = 1 \), we have already shown that the roots are \(\mathbb{F}_p\).

5. **Inductive Proof:**
   Suppose \( f_{m-1}(X) \) has roots in \(\mathbb{F}_p\). We need to show that \( f_m(X) \) has roots in \(\mathbb{F}_p\).

   Consider the polynomial:
   \[
   f_m(X) = f_{m-1}(X^p - X)
   \]
   By the inductive hypothesis, the roots of \( f_{m-1}(X) \) are in \(\mathbb{F}_p\). Therefore, the roots of \( f_m(X) \) are also in \(\mathbb{F}_p\).

6. **Conclusion:**
   By induction, the roots of \( f_m(X) \) are in \(\mathbb{F}_p\).

The final answer is \( \boxed{ \left\{ \widehat{k} \mid k \in \{0, 1, 2, \ldots, p-1\} \right\} } \) for all \( m \in \mathbb{N}^* \).