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.

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:
$$5!\cdot (44)^4$$
