Problem: Consider $n \in \mathbb{N}^+$, set $A = \mathbb{N}^+ _{\leq n}$. Find the number of monotonically increasing functions $f: A → A $ such that $f(i) \leq i$.
I tried using the multinomial theorem such that there are $x_1$ elements related to $1$, $x_2$ to $2$, and so on, using which I got $x_1+x_2+x_3+\ldots+x_n=n$. But since $x_1 \geq 1$ and $x_2,x_3,\ldots,x_n \geq 0$, we get $x_1+x_2+x_3+\ldots+x_n = n-1$. To solve this, I tried to find the coefficient of $x^{n-1}$ in $(1+x+x^2+\ldots+x^n)(1+x+x^2+x^{n-1})\ldots(1+x)$ but couldn't proceed.