Chess Board (5×5) problem 
25 small squares of a 5×5 chess board are coloured with 5 different
  colours available, such that each row contains all 5 available colours
  and no two adjacent squares have same colour. Then the no. of
  different arrangements possible are?

My attempt:
Let the colours be R,B,G,W,V
To fill the first row I have 5×4×3×2×1
Now for the second row I am sort of messing up I am able to determine 4 cases only but I am getting it at all messed up again and again.
Help me with the editing as well please.
 A: Consider the following chess table:
$$\begin{array} {|r|r|}\hline  &  &  &  &  \\ \hline  &  &  &  &  \\ \hline  &  &  &  &  \\ \hline  &  &  &  &  \\ \hline  &  &  &  &  \\ \hline  \end{array}$$
Let's fill the first row with any colors: 
$$\begin{array} {|r|r|}\hline a& b& c& d& e \\ \hline  &  &  &  &  \\ \hline  &  &  &  &  \\ \hline  &  &  &  &  \\ \hline  &  &  &  &  \\ \hline  \end{array}$$
The ways to choose $(a,b,c,d,e)$ is $5!$
For the second row, apply inclusion-exclusion.
$$\begin{array} {|r|r|}\hline a& b& c& d&e \\ \hline a'&b'  &c'  &d'  &e'  \\ \hline  &  &  &  &  \\ \hline  &  &  &  &  \\ \hline  &  &  &  &  \\ \hline  \end{array}$$
Total possible concievable arrangements of $(a',b',c',d',e')$ are again $5!$ but we must remove some unfavorable cases. 
This is known as a derangement. So, we have counted in our $5!$ some cases where one of $l = l'$ where $l$ is one of ${a,b,c,d,e}$. The number of such cases is $\binom 514!$. But, in this, we have removed some cases where $l=l'$ for two $l$ twice (once for one $l$, another time for the other). The number of such cases is $\binom 523!$. Continuing like this, we have the total good arrangements as $5!\left(\displaystyle\sum_{k = 0}^5\left(-1\right)^k\frac 1{k!}\right)$. Once we have set the arrangement for this row, we may repeat the process for the next row, and so on. 
$$\begin{array} {|r|r|}\hline a& b& c& d&e \\ \hline a'&b'  &c'  &d'  &e'  \\ \hline  a''&b''  &c''  &d''  &e''  \\ \hline  &  &  &  &  \\ \hline  &  &  &  &  \\ \hline  \end{array}$$
So, by my calculation, the total favorable arrangements is $$5!^5\left(\displaystyle\sum_{k = 0}^5\left(-1\right)^k\frac 1{k!}\right)^4$$
$$= 120*44^4\boxed{=449771520}$$
A: As you noticed for the first we have $5!$ arrangement, for the second row we can use inclusion and exclusion principle as follows.
Notably, the number of cases for the second row with at least two adjacent squares with the same colour with respect to the first row, by inclusion and exclusion principle, is:
$$5\cdot 4!-\binom{5}{2}\cdot 3!+\binom{5}{3}\cdot 2!-\binom{5}{4}\cdot 1!+\binom{5}{5}\cdot 0!=76$$
therefore the number of cases for the second row with no adjacent squares with the same colour with respect to the first row is:
$$5!-76=120-76=44$$
which is valid for all the others three rows and then finally, by multiplication rule, we obtain:
$$5!\cdot (44)^4$$
