A relatively straight forward question.
If I were to multiple any amount of different prime numbers together say 7*3*11, is it possible to divide the product by a single other prime number say 23 and have it result in an integer?
So in essence, will the product of those prime numbers only be divisible by those prime numbers and the products between those factors or will there be, for a certain multiplication of primes, a different prime that the product will be divisible by?
There is not, because the integers have the property of unique factorization, which means that each integer can be factored into primes in only one way. If an integer factors into primes $p_1p_2\ldots $ and is divisible by a prime $q$, then $q$ must be equal to one of the $p_i$.
Unique factorization is a special property of the integers. For a simple example of a system without unique factorization, consider the set $1,4,7,10,13,16\ldots$. In this system 4, 10, and 25 are prime, because we have omitted 2 and 5. The number 100 factors into the primes $4\cdot 25 $ but is also evenly divisible by the prime $10$.
For a different example, consider the set of real numbers of the form $a+b\sqrt5$, where $a$ and $b$ are integers. In this system, $4$ has two different factorizations into primes. In addition to the usual factorization $4=2\cdot 2$, one also has $4=(-1+\sqrt5)(1+\sqrt 5)$. But none of $2, -1+\sqrt5,$ and $1+\sqrt 5$ is a factor of the other two.
I advise you to read about the fundamental theorem of arithmetics. A good place to start is the wiki page about the theorem. The theorem tells you exactly that the prime divisors of a given integer are determined uniquely, meaning that $p_1^{k_1}\cdots p_n^{k_n}$ is only divisible by $p_1,\dots, p_n$.
