Greetings! I was contemplating card games, where there is initially a shuffled deck of playing cards. I decided to cycle through my deck of Penrose tiling playing cards, drawing a walk on the Cartesian plane, such that a spade corresponds to moving down, a heart corresponds to moving up, etc. Now, this created a cyclic walk of length 52 on a Cartesian grid. I then replaced the four cardinal directions (one for each suit) with 52 unit vectors (one for each card). I soon realised that the mathematics with p cards, where p is prime, is much more interesting: Associate each of the p cards with a unit vector with argument 2*pi*k/p radians (0 <= k < p). They have a sum of zero, so a walk produced by adding these vectors in any order is necessarily cyclic, returning to its original starting point. There are p! different permutations of p vectors, but (as we are only considering cycles) the start point is immaterial, so there are only (p-1)! distinct walks using this technique. Imagine the walks as (possibly self-intersecting) polygons with directed edges. It is clear that these polygons can only have order-1 or order-p rotational symmetry, as any automorphisms must preserve adjacency of vertices, and thus be cyclic permutations. The polygons with order-p rotational symmetry are the regular p-gons and star p-gons, of which there are p-1 options. There are thus (p-1)!-(p-1) walks with no (order-1) symmetry. Clearly, any rotation by a multiple of 2*pi/p radians (equivalent to rotating each of the vectors on the cards) will transform any valid walk into another. Consequently, there is an equivalence relation on the asymmetric walks, partitioning them into sets of p directly congruent walks. Hence, p divides (p-1)! - (p-1) (the number of asymmetric walks) (p-1)! - p + 1 is congruent to 0 (mod p) (p-1)! - p is congruent to -1 (mod p) (p-1)! is congruent to -1 (mod p) This is the statement of Wilson's Theorem, so the proof is complete. I suppose this is one of the rare examples where a proof of a serious mathematical theorem arises from considering something as frivolous as playing card games, where the shuffled deck imitates a random number generator. If you're (un?)lucky, I might explore PRNGs and playing cards in a future e-mail! (If it leads to proofs like this, I should play card games with fellow mathematicians in the Manchester branch of Starbucks more often... : ) Sincerely, Adam P. Goucher