The digit at the unit place of $(1!-2!+3!-\cdots+25!)^{(1!-2!+3!-\cdots+25!)}$

Question

The digit at the unit place of $(1!-2!+3!-\cdots+25!)^{(1!-2!+3!-\cdots+25!)}$ is

$a)\;0\quad b)\;1\quad c)\;5\quad d)\;9 $

My attempt: I have just tried to simplify the expression $(1!-2!+3!-\cdots+25!)$.

Now,

$(1!-2!+3!-\cdots+25!)=(1!+(3!-2!)+(5!-4!)+\cdots+(25!-24!))\\=1!+2.2!+4.4!+\cdots+24.24!$

But I am stuck here. The problem appears in ISI admission test-2010.

Answer 1

Hint. For $n\geq 5$ we have that $n!$ is divisible by 10. Hence $$N:=1!-2!+3!-\cdots+25!\equiv 1!-2!+3!-4!\equiv 1 \pmod{10}$$ that is $N$ is an integer with the unit digit equal to $1$. What is the unit digit of $N^N$?