Let $2\leq k\leq n$. Prove that $p_k(n)=p_{k-1}(n-1)+p_k(n-k)$ where $p_k(n)$ is the number of partitions of $n$ into $k$ pieces. Here's my proof:
Proof: Let $2\leq k\leq n$. Let $p_k(n)$ be the number of partitions of $n$ into $k$ parts. We can divide the partitions into two classes. First, consider all partitions that contain a part of size 1. By deleting this part, we are left with a partition of $n-1$ into $k-1$ parts. Thus there are $p_{k-1}(n-1)$ partitions in this class. Next, consider all partitions in which every part has size 2. Then by deleting 1 from every part, we are left with a partition of $n-k$ into $k$ parts. Thus there are $p_k(n-k)$ partitions in this class. Therefore, $p_k(n)=p_{k-1}(n-1+p_k(n-k)$ with the initial conditions that $p_1(n)=1$ and $p_k(n)=0$ for $n<k$.