How to calculate the probability of the following event? Assume there are $k$ identical balls, and $N$ boxes with index $1,2,\ldots,N$.
For the first round, I randomly put the $k$ balls into the $N$ boxes one by one.
Then, in the second round, I repeat the same procedure as the first round.

The question is: What is the probability of the event that the number of balls in each box is exactly the same as that of first round? Note that in each round, the empty box is allowed.
Any idea is appreciated!

An example
Thanks a lot for the comments of the question. In order to make the question more clear, I make an example. Let $N=2$ and $k=4$.
In this setting, there will be five cases for the first round: $(0,4), (1,3), (2,2), (3,1), (4,0)$. We name these five events as $E_0,E_1,E_2,E_3,E_4$. Then for the second round, the required probability is given by
\begin{align}
\sum_{i=0}^4 Pr(E_i) Pr(E_i). 
\end{align}
Now, we talk about how to calculate $Pr(E_i)$,
\begin{align}
Pr(E_i) = \frac{\frac{4!}{i!(4-i)!}}{2^4}.
\end{align}
But, for general case, can we obtain a beautiful solution?
 A: Going by @JMoravitz observation "...the random process of "putting balls randomly into boxes" assumes the balls are themselves distinct physical objects, even if they are identical to our eyes..." I am implicitly assuming that  the balls are distinct in this answer
Unless specific values are given, we can only give  the answer symbolically
The number of permutations with $k_1$ balls in box $1$, $k_2$ balls in box $2$, and so on with $0\leq k_i,$ and $\sum{k_i}=k$ is
$\dfrac {k!}{(k_1)!(k_2)!...(k_N)!}\;\;= X,$ say, and the probability of getting an identical result on repeating the experiment  will be $\dfrac{X}{N^k}$

ADDENDUM:
Can we somehow get a "proper" Pr with balls being unlabeled ?
@user2661923 has made a lot of effort in this direction.
However, leaving aside differences in interpretation as to what OP really wanted, (easily resolvable), my point is that the complicated process merely makes the balls distinct again
To explain, I'll use an Yahtzee analogue ($5$ supposedly identical balls put in $6$ distinct boxes, so that time in explaining computations is not wasted.
The four entries in each row are
group #/$\;$"Elements"$(t_i)/\;$Weightage$\,(w_i)/\quad(t_i)*(w_i)$
$G01:500000\quad\quad\; \,6\quad\quad 1\quad\quad\; 6$
$G02:410000\quad\quad 30\quad\quad \,5\quad\quad 150$
$G03:320000\quad\quad 30\quad\quad 10\quad\quad 300$
$G04:311000\quad\quad 60\quad\quad 20\quad\quad 1200$
$G05:221000\quad\quad 60\quad\quad 30\quad\quad 1800$
$G06:211100\quad\quad 60\quad\quad 60\quad\quad 3600$
$G07:111110\quad\quad 6\quad\quad 120\quad\quad 720$
........................... $252$.......................$7776$
As can be seen from the totals, $252$ corresponds to a stars and bars orientation, and $7776$ to a die throw one, so my basic point is that by using $6^5$ in the denominator, Pr computations are basically for implicitly distinct balls
A: My generic comments may have caused confusion about how I envision using Stars and Bars to attack this problem.  I think that this confusion is best cleared up with an example.
Suppose that there are $(10)$ balls and $(4)$ boxes.  Then, per Stars and Bars, there are $\binom{10 + [4-1]}{[4-1]} = (286)$ solutions that are not equiprobable.
I would partition these $(286)$ solutions into groups, where each solution (i.e. distribution of the balls into the boxes) in a specific group is equiprobable.
Group-1 (i.e. G-01)
Distribution of $(10,0,0,0)$, in some order.
Number of elements in group : $(4)$. 
Weight for each element in group: $\binom{10}{10}.$
G-02
Distribution of $(9,1,0,0)$, in some order.
Number of elements in group : $(12)$. 
Weight for each element in group: $\binom{10}{9} \times \binom{1}{1}.$
G-03
Distribution of $(8,2,0,0)$, in some order.
Number of elements in group : $(12)$. 
Weight for each element in group: $\binom{10}{8} \times \binom{2}{2}.$
G-04
Distribution of $(8,1,1,0)$, in some order.
Number of elements in group : $(12)$. 
Weight for each element in group: $\binom{10}{8} \times \binom{2}{1} \times \binom{1}{1}.$
G-05
Distribution of $(7,3,0,0)$, in some order.
Number of elements in group : $(12)$. 
Weight for each element in group: $\binom{10}{7} \times \binom{3}{3}.$
$\cdots$
So, the idea is that you would continue identifying groups, counting the number in each group, and adding a weight to each group, until you exhausted all $(286)$ possible distributions.
Now, suppose that you have $r$ groups, $G_1, G_2, \cdots G_r$.
Further suppose that group $G_r$ has $t_r$ elements, with a weight of $w_r$ given to each element in $G_r$.
Then, you will have (for example) that $\sum_{a=1}^r t_a = 286.$
Let $W = \sum_{b=1}^r \left(t_b \times w_b\right).$
Then, each of the weights should be normalized by dividing it by $W$.
Now: what are the chances that one of the specific $(286)$ configurations is repeated:
The chance that a specific configuration from G-01 is repeated is $\left(\frac{w_1}{W}\right)^2.$  Since there are $t_1$ such configurations, the chance that the distribution of balls was repeated, where the distribution is represented by some element in group G-01, is
$$t_1 \times \left(\frac{w_1}{W}\right)^2.$$
Therefore, the overall chance that the first distribution of balls into boxes is repeated on the second distribution is
$$\sum_{c=1}^r t_c \times \left(\frac{w_c}{W}\right)^2.$$

Addendum
Responding to the comments of true blue anil, following my answer:
Excerpting from the OP's question:

What is the probability of the event that the number of balls in each box is exactly
the same as that of first round?

Further, in the example given by the OP, his computation was

$\displaystyle \sum_{i=0}^4 Pr(E_i) Pr(E_i).$


Now, we talk about how to calculate $Pr(E_i)$,
$$ Pr(E_i) = \frac{\frac{4!}{i!(4-i)!}}{2^4}. $$


But, for general case, can we obtain a beautiful solution?

Also, consider the comment left by the OP following my answer:

Many thanks for your detailed answer! I think the whole logic is corret. It seems not easy to obtain the number of elements and weight in each group.

Taking your issues one at a time:
[A]
Your interpretation of the OP's question seems to be:

...that given a certain outcome of expt 1, what is the Pr that the next expt. gives an identical result.

Based on what I have excerpted, that is not the OP's intent.  That is, the OP has demonstrated that, given
a certain distribution occurring in the first placement of $k$ balls into $n$ boxes, the OP knows how to
calculate whether a specific distribution is repeated.  What the OP is actually asking is, what are the
chances that the first and second distributions will match, where it is unknown what the first distribution is.
We may need to agree to disagree on the intent of the OP's question.
Also, it can be argued that it is actually irrelevant whether the OP's intent is how you interpreted it, rather than how I interpret it.  That is, the OP can routinely use my answer (or your answer), to determine the weight $w_d$ of a distribution.  This means that the OP has demonstrated that he understands that given a certain distribution, the chance of a repeat will be
$$\left(\frac{w_d}{\text{sum of all of the weights}}\right).$$
[B]
I strongly suspect that you are right, that $W = 4^{10}.$  Whether that is true
of not, my formula for $W$ guarantees that the sum of the $(286)$ weights will
equal $W$.  This implies that my formula guarantees that $W$ is appropriate as the
normalizer of the weights.  For the
purposes of this discussion, let's assume that $W = 4^{10}.$
I agree with your math, and your perception of my intent.  Each
of the $4$ distributions in G-01 will be assigned the probability of occurring of
$\frac{45}{4^{10}}.$  I further agree that this implies that the numerator
(i.e. the probability) of each distribution occurring will vary from group to
group.  This is exactly my intent.
In reading your [B] objection, it seems that this is actually a repeat of
objection [A].  That is, if you accept my interpretation of what the OP is asking,
then it is appropriate that the "numerators" (i.e. the weights) vary from
group to group.
Another way of saying the same thing is:
If you accept my interpretation of the OP's intent, as debated in [A] above,
then I see no indication that my math is inappropriate.
