How to prove that the product of eight consecutive numbers can't be a number raised to exponent 4? How to prove this? I tried something like $$P(n,8)=\frac{n!}{(n-8)!} = b^4$$ but I can't proceed to a solution. 

Here's a proof that there is no positive integer $n$ such that
$$f(n) = (n)\cdots (n+7)$$
is a perfect $4$-th power. 

The proof is along the same lines as the proof I gave showing that there is no positive integrer $n$ such that $f(n)$ is a perfect square, but the proof for the case of $4$-th powers is easier, and can be done by hand.

Thus, suppose $n$ is a positive integer such that $f(n)$ is a perfect $4$-th power.

We start by getting a quick lower bound on $n$ . . .

Considering factors of $7$, it's clear that $f(n)$ must be a multiple of $7$, hence must be a multiple of $7^4$.

But at most two of the $8$ factors can be multiples of $7$, and at most one of them can be a multiple of $7^2$. Thus, in order for $f(n)$ to be a multiple of $7^4$, one of the $8$ factors must be a multiple of $7^3$.

It follows that $7^3 \le n + 7$, so $n \ge 336$.

Next we pair up the $8$ factors . . .

Since
\begin{align*}
n(n+7) &= n^2 + 7n\\[4pt]
(n+1)(n+6) &= n^2 + 7n + 6\\[4pt]
(n+2)(n+5) &= n^2 + 7n + 10\\[4pt]
(n+3)(n+4) &= n^2 + 7n + 12\\[4pt]
\end{align*}
it follows that
$$(n^2 + 7n)^4 < f(n) < (n^2 + 7n + 12)^4$$
hence we must have
$$f(n) = (n^2 + 7n + c)^4$$
for some positive integer $c < 12$.

From the definition of $f(n)$, we have $\bigl((n)(n+7)\bigr) \mid f(n)$.
\begin{align*}
\text{Then}\;\;&\bigl((n)(n+7)\bigr) \mid f(n)\\[4pt]
\implies\; &f(n) \equiv 0 \pmod {n^2 + 7n}\\[4pt]
\implies\; &(n^2 + 7n + c)^4 \equiv 0 \pmod {n^2 + 7n}\\[4pt]
\implies\; &c^4 \equiv 0 \pmod {n^2 + 7n}\\[4pt]
\implies\; &n^2 + 7n \le c^4\\[4pt]
\implies\; &n^2 < c^4\\[4pt]
\implies\; &n < c^2\\[4pt]
\implies\; &n < 144\qquad\text{[since $c < 12$]}\\[4pt]
\end{align*}
contrary to our lower bound, $n \ge 336$.

Hence there is no positive integer $n$ such that $f(n)$ is a perfect $4$-th power.
