counting the number of loop runs I have the following loop structure:
for $i_1=1$ to $m$
for $i_2=i_1$ to $m$
$\vdots$
for $i_n=i_{n-1}$ to $m$  
Of course, all indices $i_k$ are integers, and $m$ and $n$ are also positive integers. 
How can I count how many times the inner loop will run?

This is a problem in which it really helps to look at some small examples or to start with the simplest version and work up (or both!). The simplest version is $n=0$, which isn’t very interesting. The next simplest is $n=1$, which is also pretty trivial. The first interesting case is $n=2$, but it turns out to be helpful to have looked at some variants of the $n=1$ case first.
Suppose that there’s one instruction $I$ inside the nest of loops. If there are no loops, $I$ is executed once. If there’s one loop running from $i$ to $m$, $I$ is executed $m-i+1$ times. In particular, if there’s one loop running from $1$ to $m$, it’s executed $m$ times.
Now suppose that there are two loops. The inner loop runs from $i_1$ to $m$ for each value of $i_1$ from $1$ to $m$. Thus, $I$ is executed $m-i_1+1$ times for each value of $i_1$ from $1$ to $m$, for a total of 
$$\sum_{i_1=1}^m(m-i_1+1)=\sum_{k=1}^mk=\binom{m+1}2\;.$$
How many times is $I$ executed if the outer loop runs from $j$ to $m$ instead of from $1$ to $m$? That would be
$$\sum_{i_1=j}^m(m-i_1+1)=\sum_{k=1}^{m-j+1}k=\binom{m-j+2}2\;.$$
Now suppose that there are three loops. The second loop runs from $i_1$ to $m$ for each value of $i_1$ from $1$ to $m$, so $I$ is executed a grand total of 
$$\sum_{i_1=1}^m\binom{m-i_1+2}2\tag{1}$$ times. You’ve probably seen an identity involving binomial coefficents that will let you simplify $(1)$ to a single binomial coefficient.
You may now already be able to generalize to get the result for arbitrary $n$; if not, try repeating the argument to derive the result for $n=4$. Once you have the result, you’ll need to prove it by induction.
