Given the function \( F(t) = \prod_{i=1}^{n} \frac{t - k_i}{k_i} \), we need to show that \( F(0) = 1 \) and find \( F'(0) \).

First, we analyze \( F(0) \):

\[ F(0) = \prod_{i=1}^{n} \frac{0 - k_i}{k_i} = \prod_{i=1}^{n} \left( \frac{-k_i}{k_i} \right) = (-1)^n \]

Given that \( F(0) = 1 \), we conclude that \( (-1)^n = 1 \). This implies that \( n \) must be even.

Next, we compute the derivative \( F'(t) \). Using logarithmic differentiation:

\[ \ln F(t) = \sum_{i=1}^{n} \left( \ln(t - k_i) - \ln(k_i) \right) \]

Taking the derivative with respect to \( t \):

\[ \frac{F'(t)}{F(t)} = \sum_{i=1}^{n} \frac{1}{t - k_i} \]

At \( t = 0 \):

\[ F'(0) = F(0) \cdot \left( \sum_{i=1}^{n} \frac{1}{-k_i} \right) = (-1)^n \cdot \left( -\sum_{i=1}^{n} \frac{1}{k_i} \right) \]

Since \( n \) is even, \( (-1)^n = 1 \), thus:

\[ F'(0) = -\sum_{i=1}^{n} \frac{1}{k_i} \]

### Final Answer

F(0) = \boxed{(-1)^n}

and

F’(0) = \boxed{(-1)^{n+1} \sum_{i=1}^{n} \frac{1}{k_i}}