Let $p$ be a prime and define $f(x),g(x) \in \mathbb{Z}_p[x]$ by $f(x) = x^{p-1} - 1$ and $g(x) = (x-1)(x-2) \cdots (x-(p-1))$.
Note that the leading terms of $f(x)$ and $g(x)$ are both $x^{p-1}$, so $\deg (f(x) - g(x)) \le p-2$. Every nonzero element $\alpha \in \mathbb{Z}_p$ is a zero of both $f(x)$ and $g(x)$ (by Fermat's little theorem in the case of $f(x)$), so $\alpha$ is a zero of $f(x) - g(x)$ also. Therefore $f(x) - g(x)$ has at least $p-1$ zeroes. Because it has more zeroes than its degree, $f(x) - g(x)$ must be the zero polynomial, so $f(x) = g(x)$.
Because $f(x)$ is equal to $g(x)$, their coefficients of $x^0$ must be equal, so $-1 = (-1)(-2) \cdots (-(p-1))$ in $\mathbb{Z}_p$. This means $-1 \equiv (-1)(-2) \cdots (-(p-1)) \equiv (-1)^{p-1}(p-1)! \pmod{p}$ in $\mathbb{Z}$. If $p$ is odd then $(-1)^{p-1} = 1$, so $-1 \equiv (p-1)! \pmod{p}$ as desired. Otherwise, $p = 2$, and we use the fact that $-1 \equiv 1 \pmod{2}$.
Suppose toward a contradiction that $(n-1)! \equiv -1 \pmod{n}$ but $n$ is not prime. (To avoid triviality the problem should have stated that $n \ge 2$.) Then $n$ is composite; say $n$ is divisble by $a$ where $1 \lt a \lt n$. Therefore $(n-1)! \equiv -1 \pmod{a}$ as well. But $(n-1)! \equiv 0 \pmod{a}$ because $a$ appears in the product, so $-1 \equiv 0 \pmod{a}$, a contradiction.
Let $K$ be a field, let $\alpha_1,\ldots,\alpha_n$ be distinct elements of $K$, and let $\beta_1,\ldots,\beta_n$ be elements of $K$ (not necessarily distinct.)
First define $g_i(x) \in K[x]$ by $g_i(x) = \prod_{j \in \{1,\ldots,n\} \mathbin{\And} j \ne i} (x - \alpha_j)$. The leading term is $x^{n-1}$, so the degree is $n-1$. Clearly we have $g_i(\alpha_j) = 0$ for all $j \ne i$. Moreover, $g_i(\alpha_i)$ is a product of nonzero elements of $K$, so it is nonzero. Define $f_i(x) = g_i(\alpha_i)^{-1} \cdot g_i(x)$. Clearly we have $f_i(\alpha_j) = 0$ for all $j \ne i$, and also we have $f_i(\alpha_i) = g_i(\alpha_i)^{-1} \cdot g_i(\alpha_i) = 1$.
Define $f(x) = \beta_1 \cdot f_1(x) + \cdots + \beta_n \cdot f_n(x)$. Then $f(\alpha_i) = \beta_1 \cdot f_1(\alpha_i) + \cdots + \beta_n \cdot f_n(\alpha_i) = \beta_i$ for all $i \in \{1,\ldots,n\}$, and $f(x)$ is a sum of polynomials of degree $n-1$, so its degree is at most $n-1$.
Consider $n = 2$, $\alpha_1 = 0$, $\alpha_2 = 2$, $\beta_1 = 0$, and $\beta_2 = 1$. Suppose toward a contradiction that there is a polynomial $f(x) \in \mathbb{Z}[x]$ of degree at most $1$ (say $f(x) = a_0 + a_1x$ where $a_0,a_1 \in \mathbb{Z}$) such that $f(0) = 0$ and $f(2) = 1$. Then $a_0 = 0$ and $a_0 + a_1 \cdot 2 = 1$, so $a_1 \cdot 2 = 1$, contradicting the assumption that $a_1$ is an integer.