Odds of a number being the smallest number not picked? 
Have a game where there are $N$ players choosing some $n\in\mathbb N$,
  and the one who picked the lowest number no one else picked wins. Now,
  lets have a scenario where all players pick uniformly randomly a number from $1$ to $N$.
What are the odds that $n$ will be picked and win?
   (Talking about numbers, not players as each player is equally likely to win in this scenario)

Below are my observations so far.

Odds of no one winning are $$ \frac{\text{$f(N)$}}{N^N}$$
Where $f(N)=0, 2, 3, 40, 205\dots$ for $N=1, 2,3,4,5\dots$ players and is defined here:
 $A231797$
Now, how do we find $P(N,n)=?$ ; The odds of number $n$ winning among $N$ players who each pick some $n$ randomly from $1$ to $N$?
One trivial thing is that $P(N,n+1)\le P(N,n),\space N,n\in\mathbb N, n \le N$
 (Lower numbers win more often than larger)

We can write $P(N,n)=\frac{p(N,n)}{N^N}$, then I've found so far:
If $N<2$, then $p(N,1)=1$, else 
$$p(N,1)= N(N-1)^{(N-1)}$$
If $N \lt 3$, then $p(N,2)= 0$, else 
$$p(N,2)= p(N,1)\times \left( 1- \left(\frac{N-2}{N-1}\right)^{N-2} \right) $$
If $N \lt 3$, then $p(N,3)= 0$, else it equals $6, 36, 440, 6630, 117852\dots$  for $n=3,4,5,6,7\dots$
These so far are based on computed results:
$p(1,n)= 1$ 
$p(2,n) = 2, 0 $ 
$p(3,n) = 12, 6, 6 $ 
$p(4,n) = 108, 60, 36, 12 $ 
$p(5,n) = 1280, 740, 440, 260, 200 $ 
$p(6,n) = 18750, 11070, 6630, 3990, 2430, 1230 $ 
$p(7,n) = 326592, 195342, 117852, 71442, 43512, 26502, 17892 $ 
I've stopped at this point since I can't extract a formula for $p(N,3)$ already.


How would one mathematically solve this and find a closed expression
  for $P(N,n) ?$

 A: Let $X_n$ be the number of players choosing $n$. Then 
$$ (X_1, X_2, \ldots, X_N) \sim \text{Multinomial}\left(N; \frac {1} {N}, \frac {1} {N}, \ldots, \frac {1} {N} \right)$$
When the winning number $n = 1$, the winning condition is equivalent to $X_1 = 1$ and the probability is given by
$$ \binom {N} {1} \left(\frac {1} {N}\right)^1 \left(1 - \frac {1} {N}\right)^{N-1} = \left(1 - \frac {1} {N}\right)^{N-1}$$
which is well-mentioned in above discussion. When the winning number $n > 1$, the winning condition becomes $X_n = 1$ and $X_1 \neq 1, X_2 \neq 1, \ldots , X_{n-1} \neq 1$. Thus, the probability is 
$$ \begin{align} 
&~ \Pr\left\{(X_n = 1) \cap \bigcap_{i=1}^{n-1} (X_i \neq 1)\right\} \\ 
=&~ \Pr\left\{(X_n = 1) \cap \left(\bigcup_{i=1}^{n-1} (X_i = 1)\right)^c\right\} \\
=&~ \Pr\{X_n = 1\} - \Pr\left\{(X_n = 1) \cap \bigcup_{i=1}^{n-1} (X_i = 1)\right\} \\
=&~ \left(1 - \frac {1} {N}\right)^{N-1} - \Pr\left\{\bigcup_{i=1}^{n-1} [(X_n = 1) \cap  (X_i = 1)]\right\} \\
=&~ \left(1 - \frac {1} {N}\right)^{N-1} - \sum_{j=1}^{n-1} (-1)^{j-1}
\binom {n-1} {j} \Pr\left\{(X_n = 1) \cap \bigcap_{i=1}^{j} (X_i = 1)\right\} \\
=&~ \left(1 - \frac {1} {N}\right)^{N-1} - \sum_{j=1}^{n-1} (-1)^{j-1} 
\binom {n-1} {j} \frac {N!} {(1!)^{j+1}(N-j-1)! } \left(\frac {1} {N}\right)^{j+1} \left(1 - \frac {j + 1} {N} \right)^{N-j-1} \\
\end{align}$$
If you pull out the factor $\displaystyle \frac {1} {N^N}$, the expression becomes
$$ \frac {1} {N^N} \left[ N\left(N-1\right)^{N-1} - \sum_{j=1}^{n-1} (-1)^{j-1} 
\binom {n-1} {j} \frac {N!} {(N-j-1)!} \left(N - j - 1 \right)^{N-j-1} \right]$$
A: For $n$ to win, one player has to pick it and all the others must pick something higher.  The probability for that to happen are $N$ (to pick the player who chooses $n$) $\frac 1N$ (that the player picks $n$) $\left(\frac {N-n}N\right)^{N-1}$ (that all the other players pick larger numbers).  The winning number is most likely to be $1$, with probability $\left(\frac {N-1}N\right)^{N-1}=\left(1-\frac {1}N\right)^{N-1}$ which tends to $\frac 1e$ as $N$ gets large.  In general $n$ will be the winning number with probability about $\frac 1{e^n}$ for $n \ll N$.  The chance of having a winning number at all, again for large $N$, is about $\frac 1{e-1}$
