Partitions in Combinatorics

Question

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 $n<k$.

Answer 1

Hint

Let $P_k(n)$ be the set of partitions of $n$ with exactly $k$ parts i.e. $p_{k}(n)=|P_k(n)|$. Classify $\lambda=(\lambda_1,\dots,\lambda_k)\in P_k(n)$ (where the $\lambda_i$ are weakly decreasing and positive) based on whether $\lambda_k=1$ or $\lambda_k>1$.