Using only the definition
$$B(x, y) = \int_0^1 t^{x-1}(1-t)^{y-1}dt$$
for the Beta function $B(x, y)$, it's symmetry $B(x,y) = B(y,x)$ aswell as the fact that $(x + y)B(x + 1, y) = xB(x, y) \space\space \forall x, y > 0$ , is there a way to show that:
$$ B(m, n) = \frac{(n-1)!(m-1)!}{(n+m-1)!} \space\space\space \forall m, n \in \mathbb{N}$$
When we already presuppose other formulas and relations between the Beta function and the Gamma function which interpolates $n!$, it's probably easy (or easier at least) to argue that this is the case. But how can this be shown when only using said presumptions? Thanks in advance!
Yes, you can do it directly using integration by parts to get to $(x+y)B(x+1,y)=x B(x,y)$.
With this, you then actually want to go the other way: write the formula as $$ B(x,y) = \frac{x-1}{x+y-1} B(x-1,y). $$ Now you iterate/use induction to get to the formula $$ B(m,n) = \frac{(m-1)(m-2) \dotsm (m-(m-1))}{(m+n-1)(m+n-2)\dotsm (m+n-(m-1))} B(0,n) = \frac{(m-1)!}{(m+n-1)!/(n-2)!} B(0,n), $$ and you can do $B(0,n)$ directly to get the final $(n-1)$.