The natural sequence {$a_n$} is given. Need to prove that from some point all values of $a_n$ will be the same. Natural number $a_0$ is given. The natural sequence $\{a_n\}$ is given so that $a_n$ is the smallest possible number that satisfies this:   $$\sqrt[n]{a_0\times a_1\times a_2 \times\cdots\times a_n} = \text{natural number}.$$
Prove that after some point all the values of $a_n$ will be the same.
I've done this:
Variable $a_1$ will be $1$ because $a_0\times a_1 \in \mathbb{N}$.
If $a_0$ is a prime  then $a_0\times a_1\times a_2 = x^2$, that means $a_2 = a_0$, and in the same way $a_3 = a_0,\;\cdots,\; a_n=a_0$. That means for prime $a_0$'s it's true.
Then I tried a non-prime example for $a_0$. It was $a_0 = 2^3\times 3$.
And it really gave me that from some point $a_n$-s are the same.
I don't know how to prove this for $a_0 = p_1^{b_1}\times p_2^{b_2}\times\cdots\times p_k^{b_k}$.
 A: Hint: you can try to prove the inverse, this is, if they aren't the same at from a point forwards, then it's not going to be the smallest $a_n$, but this obvious:
We know that each prime factor $p_k$ for some $k$ needs to appear at least $n$ times, then it's clear that $p_k^n \geq 2^n$ if $p_k > 2$ for example, and for each $p_k$ we take different from all others such that we have a non-decreasing monotone sequence.
A: This may be a hint and may be a solution.
Mark the $n$-th root by $r_n$:$r_n=\sqrt[\leftroot{0}\uproot{1}n]{a_0\cdot a_1\cdot\dots\cdot a_n}$.
Note that $r_n$ is a solution for $a_{n+1}$ in sense that if you set $a_{n+1}=r_n$, then the expression $\sqrt[\leftroot{0}\uproot{1}n+1]{a_0\cdot a_1\cdot\dots\cdot a_n\cdot a_{n+1}}$ is equal to a whole (natural) number, which turns out to be exactly $r_n$ again. Now, since you know that you are looking for the smallest possible $a_{n+1}$, you know an upper bound for $a_{n+1}$, and consequently for $r_{n+1}$. What can you now tell about the sequence $\{r_n\}$? Can you compare $r_n$ and $r_{n+1}$? Later, use the fact that every monotone and bounded sequence must converge to see that your sequence also converges, or in other words, becomes constant because that's what convergence is in terms of natural numbers.
