Consider the standard Nim game, i.e. you can take as many coins as you want from a single pile, you should take at least one coin and you can't take coins from two or more different piles at the same time and last player to take a coin wins.
I am looking for a proof of the following statement:
Prove that if the number of piles are even and each pile contains two coins, the starting player loses.
The proof should be targeting an audience of people without previous exposure to combinatorial game theory or the Nim game jargon (balanced position, Nim sums etc.), and knows basic game theory. In short it should be a proof using simple (or strong) induction, nothing else.