It is known that the number of ordered $k$-tuples of positive integers who sum to $n$ is equal to $\binom{n-1}{k-1}$.
Given that result, let $p(n)$ denote the number of unordered multisets of positive integers whose sum is $n$. Show that $$p(n) \ge {\max_{1\le k\le n}} {\frac {\binom{n-1}{k-1}} {k!} }$$
Rephrasing: Given positive integers $n$ and $k$ with $n\geq k$, the number of ordered tuples of length $k$ of positive integers whose sum is $n$ is given by $\binom{n-1}{k-1}$.
This implies the number of undordered tuples of length $k$ is at least $\binom{n-1}{k-1}/k! \quad$ (here by unordered tuple we mean equivalence classes of ordered tuples, where two tuples are similar if they are permutations of each other). This is because each equivalence class has at most $k!$ elements, but can have less (when elements are repeated in a tuple).
We call the number of these unordered tuples $p_k(n)$.
Recall that the partition function $p(n)$ is defined as $\sum\limits_{k=1}^n p_k(n)$ so it follows $p(n) \geq \sum\limits_{i=1}^n\binom{n-1}{k-1}/{k!}$
Please note that "ordered tuples" are mostly just called "tuples".