The fifth and eighth terms of a geometric sequence of real numbers are $ 7!$ and $ 8!$ respectively. What is the first term?

$ \textbf{(A)}\ 60\qquad 
\textbf{(B)}\ 75\qquad 
\textbf{(C)}\ 120\qquad 
\textbf{(D)}\ 225\qquad 
\textbf{(E)}\ 315$
1. Let the first term of the geometric sequence be \( a \) and the common ratio be \( r \).
2. The \( n \)-th term of a geometric sequence is given by \( a_n = a \cdot r^{n-1} \).
3. According to the problem, the fifth term \( a_5 = 7! \) and the eighth term \( a_8 = 8! \).
4. Using the formula for the \( n \)-th term, we have:
   \[
   a_5 = a \cdot r^4 = 7!
   \]
   \[
   a_8 = a \cdot r^7 = 8!
   \]
5. We can express \( 8! \) in terms of \( 7! \):
   \[
   8! = 8 \cdot 7!
   \]
6. Substitute \( a_5 \) into the equation for \( a_8 \):
   \[
   a \cdot r^7 = 8 \cdot (a \cdot r^4)
   \]
7. Simplify the equation:
   \[
   a \cdot r^7 = 8a \cdot r^4
   \]
8. Divide both sides by \( a \cdot r^4 \) (assuming \( a \neq 0 \) and \( r \neq 0 \)):
   \[
   r^3 = 8
   \]
9. Solve for \( r \):
   \[
   r = \sqrt[3]{8} = 2
   \]
10. Now, we need to find the first term \( a \). Using \( a_5 = a \cdot r^4 = 7! \):
    \[
    a \cdot 2^4 = 7!
    \]
11. Simplify:
    \[
    a \cdot 16 = 7!
    \]
12. Solve for \( a \):
    \[
    a = \frac{7!}{16}
    \]
13. Calculate \( 7! \):
    \[
    7! = 5040
    \]
14. Substitute \( 7! \) into the equation:
    \[
    a = \frac{5040}{16} = 315
    \]

The final answer is \( \boxed{315} \).