HINT
If $p$ is a prime, then $p$ divides $\dbinom{p}{k}$ for all $k \in \{1,2,\ldots,p-1\}$.
The identity you have i.e. $(x+y)^p = x^p + y^p$, is referred to as Freshman's dream.
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Note that $\dbinom{p}{k} = \dfrac{p \times (p-1) \times (p-2) \times \cdots \times (p-k+2) \times (p-k+1)}{k \times (k-1) \times (k-2) \times \cdots \cdots \times 2 \times 1}$ is an integer. Since $k \in \{1,2,\ldots,p-1\}$, and $p$ is a prime, none of $k, k-1, k-2, \ldots, 2$ divide $p$. Hence, we can factor $p$ from the numerator to get that $\dbinom{p}{k} = \dfrac{p \times (p-1) \times (p-2) \times \cdots \times (p-k+2) \times (p-k+1)}{k \times (k-1) \times (k-2) \times \cdots \cdots \times 2 \times 1} = p \times M$ where $M \in \mathbb{Z}$.