Q:

Probability of double matchings

Yesterday, I solved this problem question from probability theory: Two similar decks of $N$ distinct cards are matched against a similar target deck. Find the probability of exactly $m \leq N$ matches.
I proceeded in the following manner. Let $A_i$ denote the event that $i^{\text{th}}$ card is matched (from both the decks) against the target deck. Therefore, let $$P(\text{E = exactly $m$ match occurs})(N,m)$$ (don't mind the bad notation please) then
$$P(N,m) = S_m - \binom{m+1}{m}S_{m+1} + \binom{m+2}{m}S_{m+2} + \ldots + \binom{N}{m} S_{N}$$
where $S_1 = \sum_{1\leq i \leq N} P(A_i)$, $S_2 = \sum_{1\leq i \lt j \leq N} P(A_i \cap A_j) \ldots$
Clearly, we have $$S_{m+k} = \binom{N}{m+k} \frac{(N-m-k)!^2}{N!^2}$$
Therefore, 
\begin{align*}
P(N,m) &= \sum_{k=0}^{N-m} (-1)^k \binom{m+k}{m} \binom{N}{m+k} \frac{(N-m-k)!^2}{N!^2} \\ 
&= \frac{1}{m!} \frac{1}{N!} \sum_{k=0}^{N-m} (-1)^k \frac{(N-m-k)!}{k!}
\end{align*}
After obtaining the above expression, I thought if there exists some nice closed formula for the series. So I plugged it on W|A but it doesn't returns one (in terms of elementary functions).
Next, I started wondering, how does this probability function behaves as $N \rightarrow \infty$. Because this limit might be actual translation of some real world phenomena (although that is something to be considered about, later on).
So, I first tried to check for $m=0$
\begin{align*}
\lim_{N \rightarrow \infty} P(N,0) &= \lim_{N \rightarrow \infty} \frac{1}{N!} \sum_{k=0}^{N} (-1)^k \frac{(N-k)!}{k!} \\
&= \lim_{N \rightarrow \infty} \left(1 - \frac{1}{N} + \frac{1}{2!}\frac{1}{N(N-1)} - \ldots \right)
\end{align*}
It doesn't strikes me on how to solve this limit, as I cannot evaluate the limit pointwise since it is an infinite sum. So I thought of setting up a recurrence (which may help?). This is what I found:
$$P(N+2,0) = P(N+1,0) + \frac{P(N,0)}{(N+1)(N+2)} + \frac{(-1)^{N}}{(N+2)!^2}$$
But again, I still couldn't figure out much. I even expressed this as an integral (just because sometimes, it does help) and then tried to do some manipulations, but still no clue
$$P(N,0) = (N+1) \int_0^1 \sum_{k=0}^N\frac{ t^k(1-t)^{N-k}}{k!^2} \mathrm{d}t$$ 
So these are the questions, I am trying to find a solution to:

Is there any nice closed form for the expression?
How does the probability function, which I derived, behaves when $N \rightarrow \infty$ for a fixed $m$?
What happens as $N \rightarrow \infty$ and $m \rightarrow \infty$?

Any help would be appreciated.
Edit 1: I figured out that $P(N,0) \rightarrow 1$ as $n \rightarrow \infty$ with some computations but I guess, it still requires a rigorous proof.

A:

Let $X_k = \mathbf{1}_{A_k}$, $k=1,\dots,N$, be the variable indicating a (double) match in the $k$th card so that $X = X_1+\dots + X_N$ is the number of matches. By linearity, 
$$
\mathbb{E}[X] = \sum_{k=1}^N \mathbb{E}[X_k] = N\cdot \frac{1}{N^2} = \frac{1}{N}.
$$
By the Markov inequality, $\mathbb{P}(X\ge 1) \le \mathbb{E}[X] \to 0$, $N\to\infty$. Therefore, $P(N,0)\to 1$, $N\to\infty$.

