Consider the classical Nim game with total number of stones being odd. Then the first players wins, of course, what follows from the general description of winning positions. But is there some shorter (independent of full theory) explanation of this fact, maybe with implicit strategy or whatever?



Nim game for odd number of stones
The following strategy-stealing argument is similar to the one in my answer to ["An unfair game involving an odd number of pieces of chocolate."](https://mathoverflow.net/questions/127116/an-unfair-game-involving-an-odd-number-of-pieces-of-chocolate)


Inductively assume that all positions with a smaller odd number of stones are a win for the first player. 


Consider taking one stone from a pile of odd size. If this is a winning play, we are done. If not, then your opponent has a winning response. This must not be an odd number of stones by induction. To have a chance of winning, your opponent must take an even number of stones. Now, try to take an even number of stones as long as possible, and your opponent's winning strategy must be to play even moves until you make another odd play. En route to winning, your opponent has to force you to take an odd number of stones while only taking even amounts. 


To create a winning strategy, you steal this response to taking one stone from an odd pile. If your opponent took $2n$ from an originally even pile as a response to your first move, then it must be a winning first play for you to take $2n-1$ from that pile, then copy the strategy for forcing you to take an odd number of stones using only even moves. If your opponent took $2n$ from an originally odd pile (including the one from which you took a stone) as a response to your first move, then it must be a winning play for you to take $2n+1$ from that pile, then copy the strategy. This means you can leave an even number of stones after each move, and you can force your opponent to take an odd number of stones at some point. By induction, this gives you a winning position. 


This theft works because you never change the greatest even integer smaller than each heap size. Every even move available to your opponent after you took one stone from an odd pile is available to you, and there are no extra even moves available for your opponent.
$\square$


While this argument didn't require an understanding of the solution to Nim, it is easy to see what is going on when you understand that the second player wins are positions with balanced Nim sums. If you take one stone from an odd pile, you balance the ones digit without changing anything else. Your opponent's winning play, if any, tells you how to balance the other digits. A first play which takes one more or one less stone depending on the original parity of that heap balances all digits. For example, from $\star15 \star12 \star5 \star 3,$ suppose you take $1$ from $\star5$. Your opponent balances the position by taking $4$ from $\star15$, $\star12$, or $\star4$. This tells you that in the original position, taking $5$ from $\star15$ or $\star 5$, or taking $3$ from $\star 12$ will create a balanced position.


