Which would be the "direct" formula for the birthday paradox , withouth subtracting to 1?

Probably the simplest approach is to consider how many days are somebody's birthday. If this is $b$, you want to add up the probabilities for $1\le b \le n-1$ which with $d=365$ is $$\sum_{b=1}^{n-1} \frac{d! \,S_2(n,b)}{(d-b)! \,d^n} $$ where $S_2(n,b)$ represents Stirling numbers of the second kind. This gives the same answer as more conventional $1-\frac{d!}{(d-n)! \,d^n}$

For example, if $n=5$, this would suggest

$$\frac{365\times 1 }{365^5}+\frac{ 365\times 364\times15 }{365^5}+\frac{ 365\times 364\times363\times 25 }{365^5}+\frac{365\times 364\times363\times362\times 10 }{365^5}$$

which gives the same answer as $1 - \frac{365}{365}\times \frac{364}{365}\times \frac{363}{365}\times \frac{362}{365}\times \frac{361}{365}$, both being $\frac{175793709365}{6478348728125}\approx 0.027$