How many onto functions are there from a set $|T|=n+1$ to a set $|S|=n$? How many onto functions are there from a set $|T|=n+1$ to a set $|S|=n$? 
It is well known that the number of onto functions from $|T|=m$ to $|S|=n$ is given by $$  n!\left(m\brace n\right)$$ where $m\brace n$ denotes a Sterling number of the second kind. 
However, when $m=n+1$, I suppose the the computation may be simpler and not require this formula. 
Thus we want to prove the number of surjections from $T\to S$, with $|T|=n+1$ and $|S|=n$ is $$\frac{n(n+1)!}{2}=\frac{n(n+1)}{2}\cdot n!=(1+...+n)n~!$$
Consider the simpler case from here by momentarily letting $T^*=T\setminus\{n+1\}$. Thus, $|T^*|=n$. 
We find the number of onto functions from $T^*\to S$. From here it is easy to see that there are $n!$ such functions since the first element of $T^*$ has a choice of $n$ elements to map to in $S$, then the second element in $T$ of has a choice of $n-1$ elements to map to in $S$, ... , and so on. Thus there are $$n\cdot(n-1)\cdot(n-2)\cdot...\cdot1=n!$$
such surjections.
Consider $T^*\cup\{n+1\}=T\therefore|T^*\cup\{n+1\}|=n+1$.
Since there are $n!$ onto functions that map $\{1,...,n\}\to\{1,...,n\}$, an onto map from $\{1,...,n,n+1\}\to\{1,...,n\}$ must map two $n_T\in\{1,...,n,n+1\}$ into the same element in $\{1,...,n\}$, thus there are at least $$n\cdot n!$$ such onto functions now. 
Here is where I am stuck. How do I count what that last added element adds to the number of surjections?

An onto function from a set with $n+1$ elements to a set with $n$ elements will necessarily have $2$ elements from the domain mapping to the same element in the codomain and then all remaining elements in the domain mapping to a unique element in the codomain.


*

*Pick which two elements from the domain share an output

*Pick what that output was

*From smallest to largest, pick what the output is for each remaining element
The number of such functions is then:
$$\binom{n+1}{2}n(n-1)! = (n+1)!\cdot \frac{n}{2}$$
