Show that any integer $n > 1$ can be written as $n = p_1 \times p_2 × \ldots × p_r × N^2$ where $p$’s are distinct primes.
By the fundamental theorem of arithmetic, any integer larger than 1 can be written uniquely as a product of primes.
$$ n = p_1^{k_1} \times p_2^{k_2} \times \ldots \times p_x^{k_x}$$
Each prime factors $p_i$ appears $k_i$ times in this product.
There are two cases for $k_i$, even $k_i=2m$ and odd $k_i=2m+1$, for some positive number $m$.
Case $k_i$ even
We can write each prime factor that appears an even number of times as
$$ p_i^{k_i} = p_i^{2m} = (p_i^m)^2 $$
And so each $p_i$, which appears an even number of times in the prime factorisation of $n$, is a factor of $N^2$.
Case $k_i$ odd
Similarly, we can write each prime factor that appears an odd number of times as
$$ p_i^{k_i} = p_i^{2m+1} = p_i \times (p_i^m)^2 $$
And so each $p_i$ which appears an odd number of times in the prime factorisation of $n$, appears once in the product of unique primes, and the remaining even number of $p_i$ are factors of $N^2$.
And so any integer $n>1$ can be written as a product of distinct primes and a perfect square $N^2$.