Here is the problem I am trying to solve:
Let $n \in \mathbb N = \{1,2, \dots\}, n = p_1^{a_1} \dots p_l^{a_l}, \, a_i \geq 1 \, , \, p_1 < p_2 < \dots ,$ so $\operatorname{rad}(n) = (p_1 \dots p_l) \,.$
Where $\operatorname{rad}(n)$ is the radical of the ideal $(n)$ which is defined as $$\{x \in A: x^k \in (n) \textbf{ for some } k > 0\}.$$
I do not know how to tackle this problem actually, I know how to tackle it if $n = 2 ^k$ but in the above general case I do not know what to do.
In case of $n = 2^k,$ we know that $\operatorname{rad}(n) = 2$ so we proceeded by showing that if you have a radical, then you have an even number and conversely if it is a radical then it is an even number. But this prove depended on knowing the prime itself.
Could someone help me please?
From the problem statement, you know that the answer should be rad$(n) = (p_1\cdots p_l)$ (by the way, no commas inside the RHS, we must multiply all of the primes together). This means we must show that given $m \in \Bbb{Z}$, we have $n | m^k$ for some $k \geq 0$ iff $p_1,\cdots,p_l | m$. I think the simplest approach is to just follow through the definitions.
Suppose $m \in \text{rad}(n)$. Then $n|m^k$ for some $k\geq 0$. Since $p_i | n$ for each $1 \leq i \leq l$, we have $p_i | m^k$, hence $p_i | m$ by Euclid's Lemma. As the $p_i$ are coprime, we have $p_1\cdots p_l | m$. Therefore $m \in (p_1\cdots p_l)$.
Now suppose $m \in (p_1\cdots p_l)$. Then $m = cp_1\cdots p_l$ for some $c \in \Bbb{Z}$. Let $a = \max\{a_1, \cdots, a_l\}$. Then we see $$ m^a = (cp_1\cdots p_l)^a = c^ap_1^a\cdots p_l^a = \left(c^ap_1^{a-a_1}\cdots p_l^{a-a_l}\right)\cdot\left(p_1^{a_1}\cdots p_l^{a_l}\right) = \left(c^ap_1^{a-a_1}\cdots p_l^{a-a_l}\right)n. $$ Thus $n | m^a$ and so $m \in \text{rad}(n)$ as desired.
There is a reason why the "radical" is called what it is. You might of noticed that taking the radical of $(n)$ essentially killed the powers on the prime factors of $n$. This is because we have to "power up" to get inside of rad$(n)$. This result is not unique to the integers either--it applies to any nonzero proper principal ideal in any unique factorization domain (UFD).