Let $f(x)=\cos(\cos(\cos(\cos(\cos(\cos(\cos(\cos(x))))))))$, and suppose that the number $a$ satisfies the equation $a=\cos a$. Express $f'(a)$ as a polynomial in $a$.
1. **Define the sequence of functions**: 
   Let \( f_0(x) = x \) and \( f_n(x) = \cos(f_{n-1}(x)) \). We are given \( f_8(x) \).

2. **Induction hypothesis**: 
   We will prove by induction that \( f'_n(x) = \prod_{k=0}^{n-1} (-\sin(f_k(x))) \).

3. **Base case**: 
   For \( n = 1 \),
   \[
   f_1(x) = \cos(x) \implies f'_1(x) = \frac{d}{dx} \cos(x) = -\sin(x) = -\sin(f_0(x))
   \]
   This matches our hypothesis for \( n = 1 \).

4. **Inductive step**: 
   Assume \( f'_n(x) = \prod_{k=0}^{n-1} (-\sin(f_k(x))) \) holds for some \( n \). We need to show it holds for \( n+1 \).
   \[
   f_{n+1}(x) = \cos(f_n(x)) \implies f'_{n+1}(x) = \frac{d}{dx} \cos(f_n(x)) = -\sin(f_n(x)) \cdot f'_n(x)
   \]
   By the chain rule, substituting the induction hypothesis,
   \[
   f'_{n+1}(x) = -\sin(f_n(x)) \cdot \prod_{k=0}^{n-1} (-\sin(f_k(x))) = \prod_{k=0}^{n} (-\sin(f_k(x)))
   \]
   This completes the induction.

5. **Evaluate at \( a \)**: 
   Given \( a = \cos(a) \), it follows that \( f_n(a) = a \) for all \( n \). Therefore,
   \[
   f'_8(a) = \prod_{k=0}^{7} (-\sin(f_k(a))) = \prod_{k=0}^{7} (-\sin(a)) = (-\sin(a))^8 = \sin^8(a)
   \]

6. **Express \(\sin^8(a)\) in terms of \( a \)**: 
   Using the Pythagorean identity, \( \sin^2(a) + \cos^2(a) = 1 \), and since \( a = \cos(a) \),
   \[
   \sin^2(a) = 1 - a^2
   \]
   Therefore,
   \[
   \sin^8(a) = (\sin^2(a))^4 = (1 - a^2)^4
   \]

7. **Expand \((1 - a^2)^4\)**: 
   Using the binomial theorem,
   \[
   (1 - a^2)^4 = \sum_{k=0}^{4} \binom{4}{k} (-a^2)^k = 1 - 4a^2 + 6a^4 - 4a^6 + a^8
   \]

Thus, 
\[
f'_8(a) = 1 - 4a^2 + 6a^4 - 4a^6 + a^8
\]

The final answer is \(\boxed{a^8 - 4a^6 + 6a^4 - 4a^2 + 1}\)