Ten discs on three pegs

Idea. The largest disc must move at least once, and when it moves, all the other discs must sit on the third peg. So the $n-1$ smaller discs must be moved out of the way, then back on top: twice the smaller problem, plus one move.

Step 1: the recursion. To move the largest disc from peg A to peg C, the $n-1$ smaller discs must all be on peg B. Getting them there takes at least $T_{n-1}$ moves, and putting them back on top of the largest disc takes at least $T_{n-1}$ more. So $T_n\ge 2T_{n-1}+1$, and the obvious recursive strategy achieves it: $$T_n=2T_{n-1}+1,T_1=1.$$

Step 2: solve. $T_n+1=2(T_{n-1}+1)$, so $T_n+1=2^n$ and $T_n=2^n-1$.

Step 3. For 10 discs, $T_{10}=2^{10}-1=1023$.

The legend. The puzzle's story has 64 golden discs: $2^{64}-1$ moves, about 585 billion years at one move per second.