Well, you can certainly compute $p(n+1)/p(n)$. But it's not a trivial expression. Let $q(n)=1-p(n)$. Then we know that $$q(n+1)=q(n)\left(1-\frac{n}{365}\right)$$ So $$\begin{align}p(n+1) = 1-q(n+1) &= 1-q(n)\left(1-\frac{n}{365}\right)\\& = 1-(1-p(n))\left(1-\frac{n}{365}\right)\\=p(n) +\frac{(1-p(n))n}{365}\end{align}$$

So $\dfrac{p(n+1)}{p(n)}$ is going to be a messy expression of $p(n)$.

Similarly $p(n+1)-p(n)=\frac{(1-p(n))n}{365}$.

Essentially, this means that we get a matching birthday in $n+1$ people if we get a matching birthday in the first $n$, or if we don't in the first $n$ and the $n+1$st matches one of the $n$ previous ones.

Well, you can certainly compute $p(n+1)/p(n)$. But it's not a trivial expression. Let $q(n)=1-p(n)$. Then we know that $$q(n+1)=q(n)\left(1-\frac{n}{365}\right)$$ So $$\begin{align}p(n+1) = 1-q(n+1) &= 1-q(n)\left(1-\frac{n}{365}\right)\\& = 1-(1-p(n))\left(1-\frac{n}{365}\right)\\=p(n) +\frac{(1-p(n))n}{365}\end{align}$$

So $\dfrac{p(n+1)}{p(n)}$ is going to be a messy expression of $p(n)$.

Similarly $p(n+1)-p(n)=\frac{(1-p(n))n}{365}$.

Essentially, this means that we get a matching birthday in $n+1$ people if we get a matching birthday in the first $n$, or if we don't in the first $n$ and the $n+1$st matches one of the $n$ previous ones.

For the example of $n=3$:

$$\begin{align}p(1)&=0\\p(2)&=p(1)+\frac{1(1-p(1))}{365} = \frac{1}{365}\\p(3)&=p(2)+\frac{2(1-p(2))}{365}=\frac{1}{365} + \frac{2\cdot 364}{365^2}\approx 0.0082042\end{align}$$

Well, you can certainly compute $p(n+1)/p(n)$. But it's not a trivial expression. Let $q(n)=1-p(n)$. Then we know that $$q(n+1)=q(n)\left(1-\frac{n}{365}\right)$$ So $$\begin{align}p(n+1) = 1-q(n+1) &= 1-q(n)\left(1-\frac{n}{365}\right)\\& = 1-(1-p(n))\left(1-\frac{n}{365}\right)\\=p(n) +\frac{(1-p(n))n}{365}\end{align}$$

So $\dfrac{p(n+1)}{p(n)}$ is going to be a messy expression of $p(n)$.

Similarly $p(n+1)-p(n)=\frac{(1-p(n))n}{365}$.

You might be able to give an argument for the latter.

Well, you can certainly compute $p(n+1)/p(n)$. But it's not a trivial expression. Let $q(n)=1-p(n)$. Then we know that $$q(n+1)=q(n)\left(1-\frac{n}{365}\right)$$ So $$\begin{align}p(n+1) = 1-q(n+1) &= 1-q(n)\left(1-\frac{n}{365}\right)\\& = 1-(1-p(n))\left(1-\frac{n}{365}\right)\\=p(n) +\frac{(1-p(n))n}{365}\end{align}$$

So $\dfrac{p(n+1)}{p(n)}$ is going to be a messy expression of $p(n)$.

Similarly $p(n+1)-p(n)=\frac{(1-p(n))n}{365}$.

Essentially, this means that we get a matching birthday in $n+1$ people if we get a matching birthday in the first $n$, or if we don't in the first $n$ and the $n+1$st matches one of the $n$ previous ones.