Consider the integer $N = (11 \cdot 31 \cdot 41 \cdot 61 \cdots p_n)^5 - 1$. Then $N$ is relatively prime to $11, 41, \cdots p_n$. Let $a = 11 \cdot 31 \cdot 41 \cdot 61 \cdots p_n$, then
$$N = a^5 - 1 = (a-1)(a^4 + a^3 + a^2 + a + 1).$$
Clearly, $2 \nmid (a^4 + a^3 + a^2 + a + 1)$, but $5 \nmid (a^4 + a^3 + a^2 + a + 1)$. Let $p \neq 5$ be a prime factor of $a^4 + a^3 + a^2 + a + 1$. Therefore, $p \nmid (a-1)$. The reason is as follows:

If $p \nmid (a-1)$, then $a = kp + 1$, for some integer $k$, which gives
$$a^2 = (kp + 1)^2, \quad a^3 = (kp + 1)^3, \quad a^4 = (kp + 1)^4,$$
and
$$a^4 + a^3 + a^2 + a + 1 = (kp+1)^4 + (kp+1)^3 + (kp+1)^2 + (kp+1) + 1 \equiv 5 \pmod{p}.$$
In fact, $(p-1) \equiv 4 \pmod{5}$, that is, $p-1 = 5k + 4$.

By Fermat's theorem, $p \nmid (a^{p-1} - 1)$. But at this point
$$a^{p-1} - 1 = a^{5k+4} - 1 = a^4(a^{5k} - 1) + (a^4 - 1),$$
and since $(a^5 - 1) \nmid (a^{5k} - 1) = (a^5)^k - 1^k$, this means $p \nmid (a^{5k} - 1)$. That is, $p \nmid (a^4 - 1)$. However
$$a^5 - 1 = a(a^4 - 1) + (a - 1)$$
Therefore, if $p \nmid (a^5 - 1)$ and $p \nmid (a^4 - 1)$, then $p \nmid (a - 1)$. The above cannot happen. By a similar approach, we obtain that the remainder of $(p-1)$ divided by $5$ cannot be $1, 2, 3$.

Hence $5|(p-1)$ and $p-1$ is even. We get $10|(p-1)$, that is, $p = 10k + 1$, so $p$ is one of the terms of the given arithmetic sequence. Therefore, the prime factors of $a^4 + a^3 + a^2 + a + 1$ are $5$ and primes of the form $10k + 1$.

However $a^4 + a^3 + a^2 + a + 1 > 5$, and $5^2 \nmid (a^4 + a^3 + a^2 + a + 1)$. In fact, the integer $a$ has last digit $1$, that is, $a = 5k + 1$. By the binomial theorem, we get
$$\begin{aligned} a^4 + a^3 + a^2 + a + 1 &= (5k + 1)^4 + (5k + 1)^3 + (5k + 1)^2 + (5k + 1) + 1 \\ &= 5 \cdot [5(25k^4 + 25k^3 + 10k^2 + 2k) + 1]. \end{aligned}$$
Therefore $N = a^5 - 1$ has at least one prime factor of the form $10^k + 1$. But by the above, $N$ is relatively prime to all numbers of the form $10^k + 1$. This is a contradiction.