Prove that $N – \lfloor{N/p}\rfloor = \lfloor{\frac{p-1}{p}\left({N + 1}\right)}\rfloor$ for positive $N$ and prime $p$

I am counting the number of positive integers less than or equal to some positive integer $ N$ and not divisible by some prime $ p$ . This gets generalized for $ k$ primes where I use the principle of inclusion-exclusion for this result. The simple result for a single prime is $ N – \lfloor{\frac{N}{p}}\rfloor$ . However, I have noticed by experiment that this is also equal to $ \lfloor{\frac{P-1}{p} \left({N + 1}\right)}\rfloor$ . I am looking for a proof of this relation if true.