Find the remainder when 13⁴⁰ is divided by 14, and the remainder when 13⁴¹ is divided by 14.

$(d + 1)^n$ leaves remainder $1$ on division by $d$, and $(d - 1)^n$ leaves $1$ for even $n$ and $d - 1$ for odd $n$. Modulo $d^2$, $(d + 1)^n \equiv 1 + nd$. To prove divisibility, write each power as $(\text{multiple} \pm 1)^n$ and expand. For small $h$, $(1 + h)^n \approx 1 + nh + {}^{n}C_{2}h^2$.