About putting $n$ distinct balls into $n$ distinct boxes. Let the balls be labelled $1,2,3,..n$ and the boxes be labelled $1,2,3,..,n$. 
Now I want to find, 


*

*What is the expected value of the minimum value of the label among the boxes which are non-empty 

*What is the expected number of boxes with exactly one ball in them? 

Whatever way I am thinking of it, I am getting complicated summation form of answers and not any exact closed form! 
 A: A
To expectation of the minimum used label, $K$, we first measure the probability that all the balls being among the top $n-k$ boxes.  That is, that the minimum label will be greater than some value $k$.
In the total space each of $n$ balls has a choice of $n$ boxes ($n^n$). In the restricted space each of $n$ balls has a choice of $n-k$ boxes $(n-k)^n$.  So then:
$$\begin{align}
\Pr(K > k) & = \frac{(n-k)^n}{n^n}
\\[1ex] \Pr(K > k-1) & = \frac{(n-k+1)^n}{n^n}
\\[1ex] \Pr(K=k) & = \Pr(K>k-1)-\Pr(K>k)
\\ & = \frac{(n-k+1)^n-(n-k)^n}{n^n}
\\[2ex]
E(K) & = \sum_{k=1}^{n} k \Pr(K=k)
\\ & = \frac 1 {n^n}\sum_{k=1}^n (k(n-k+1)^n-k(n-k)^n)
\\ & = \frac 1 {n^n}\left(\sum_{k=0}^{n-1} (k+1)(n-k)^n-\sum_{k=1}^{n} k(n-k)^n\right)
\\ & = \frac 1 {n^n}\left( n^n + \sum_{k=1}^{n-1} (n-k)^n\right)
\\ & = \frac 1 {n^n}\left( n^n + \sum_{k=1}^{n-1} k^n\right)
\\ & = \mathop{1 + n^{-n}\underbrace{\sum_{k=1}^{n-1} k^n}}_{\text{a generalised Harmonic series?}}
\end{align}$$

B
Let $B_i$ be the Bernouli indicator that box $i$ contains exactly one ball.   We then use the linearity of expectation to find the expected value of $B:=\sum_{i=1}^n B_i$
$$\begin{align}
\forall i\in \{1..n\}:\Pr(B_i=1) & = {n\choose 1}(n-1)^{n-1}/n^n
\\[2ex] \mathbb{E}[B] & = \mathbb{E}[\sum_{i=1}^n B_i]
\\ & = \sum_{i=1}^n \mathbb{E}[B_i]
\\ & = n\times (0\times \Pr(B_\ast=0)+1\times \Pr(B_\ast=1))
\\[1ex]\therefore \mathbb{E}[B] & = \frac{(n-1)^{n-1} }{ n^{n-2}}
\end{align}$$
A: Here is an approach  using labelled species and exponential generating
functions.

For the first problem we have the species
$$\sum_{q=1}^n \mathcal{U}^q \times 
\mathfrak{P}_{\ge 1}(\mathcal{Z}) \times 
\mathfrak{P}(\mathcal{Z})^{n-q}$$
with  $\mathcal{U}$ marking  the end  of the  intial segment  of empty
bins.
This yields the generating function
$$\sum_{q=1}^n u^q (\exp(z)-1) \exp(z)^{n-q}
= (\exp(z)-1) \exp(z)^n \sum_{q=1}^n u^q \exp(z)^{-q}.$$
Some algebra produces
$$(\exp(z)-1) \exp(z)^{n-1} \times u
\frac{(u/\exp(z))^n - 1}{u/\exp(z)-1}.$$
which is
$$(\exp(z)-1)
\frac{u^{n+1} - u \exp(z)^n}{u-\exp(z)}.$$
Differentiate and put $u=1$ to obtain the EGF of the count
$$\left.(\exp(z)-1)
\left(\frac{(n+1)u^n - \exp(z)^n}{u-\exp(z)}
- \frac{u^{n+1} - u \exp(z)^n}{(u-\exp(z))^2}
\right)\right|_{u=1}
\\= (\exp(z)-1)
\left(\frac{(n+1) - \exp(z)^n}{1-\exp(z)}
- \frac{1 - \exp(z)^n}{(1-\exp(z))^2}
\right)
\\ = \frac{1 - \exp(z)^n}{1-\exp(z)}
+ \exp(z)^n - (n+1)
= -(n+1) + \frac{1 - \exp(z)^{n+1}}{1-\exp(z)}.$$
Performing coefficient extraction we obtain for $n\ge 1$ the answer
$$\frac{1}{n^n} n! [z^n]
\left(-(n+1) + \frac{1 - \exp(z)^{n+1}}{1-\exp(z)}\right)
= \frac{1}{n^n} n! [z^n]
\left(-(n+1) + \sum_{q=0}^n \exp(z)^q\right)
\\ = \frac{1}{n^n} n! \sum_{q=1}^n \frac{q^n}{n!}
=  \frac{1}{n^n} \sum_{q=1}^n q^n
= 1 + \frac{1}{n^n} \sum_{q=1}^{n-1} q^n.$$
For the second problem the species is
$$\mathfrak{S}_{=n}
\left(\mathcal{U}\mathcal{Z}+\mathfrak{P}_{\ne 1}(\mathcal{Z})\right).$$
with $\mathcal{U}$ marking singletons.

This gives the generating function
$$\left(uz + \exp(z) - z\right)^n.$$
For the  expected number of  singletons differentiate with  respect to
$u$ and set $u=1$ to obtain 
$$\left. n \left(uz + \exp(z) - z\right)^{n-1} \times z\right|_{u=1}
= n z \exp(z)^{n-1}.$$
Finally extract coefficients for the expectation which is
$$\frac{1}{n^n} n! [z^n] n z \exp(z)^{n-1}
= \frac{1}{n^{n-1}} n! [z^{n-1}] \exp(z)^{n-1}
= \frac{1}{n^{n-1}} n! \frac{(n-1)^{n-1}}{(n-1)!}
\\= \frac{1}{n^{n-1}} n (n-1)^{n-1}
= \frac{(n-1)^{n-1}}{n^{n-2}}.$$
These results match the first answer.
Remark. Since
$$\frac{(n-1)^{n-1}}{n^{n-2}} = (n-1)\left(1-\frac{1}{n}\right)^{n-2}
= (n-1)\left(1-\frac{1}{n}\right)^n \left(1-\frac{1}{n}\right)^{-2}
\\ = \frac{n^2}{n-1} \left(1-\frac{1}{n}\right)^n$$
the second expectation is
$$\frac{1}{e} \frac{n^2}{n-1}.$$
