Let's simplify your question first. What you have is:
$$\sum_{j=0}^n \left(-\sum_{t=0}^k f(k,t) g(j,t)\right)$$
for some function $f$ (in your particular case: $f(k,t) = {k+1\choose t}(-1)^{k+1-t}$ and $g$ (in your case, $g(j, t) = j^t$.
Let's forget what $g$ and $g$ are for the moment because it's not relevant.
The sum $$S=\sum_{j=0}^n \left(-\sum_{t=0}^k f(k,t) g(j,t)\right)$$
can be first rewritten as
$$S=-\sum_{j=0}^n \left(\sum_{t=0}^k f(k,t) g(j,t)\right)$$
Now, let's reverse the order of summing. In the original case, you sum something from $j=0$ to $j=n$, and for every $j$, you sum the inner sum from $t=0$ to $t=k$, which means you cover the whole square $\{(j,t)\in\mathbb Z^2| 0\leq j\leq n\land 0\leq t\leq k\}$.
You can cover the same square by summing first over $t$ and then over $j$, so you get
$$S=-\sum_{t=0}^k \left(\sum_{j=0}^n f(k,t) g(j,t)\right)$$
Now, for every value of $t$, you can see that
$$\sum_{j=0}^n f(k, t)g(j,t) = f(k,t)\sum_{j=0}^n g(j,t)$$
because you are just factoring out the common factor $f(k,t)$. It becomes ogvious if you write it as
$$\sum_{j=0}^n f(k, t)g(j,t) = f(k,t) g(0,t) + f(k,t)g(1,t) + \dots + f(k,t)g(n,t) = f(k,t) (g(0,t) + g(1,t) + \dots + g(n,t)) = f(k,t)\sum_{j=0}^n g(j,t)$$
So now you get
$$S=-\sum_{t=0}^k \left(f(k,t)\sum_{j=0}^n g(j,t)\right)$$
which is what you wanted to have in the first place.
First use some notation to define your sample. Let $X_1, X_2,\ldots,X_n$ be the $n$ members in your sample. If $f$ is the function you're applying to each member, then the sum you are looking for can be written
$
\sum_{i=1}^n f(X_i)
$.
If you have a different set of $X$'s within each outer iteration, then you could modify your notation: $X_{1,1}, X_{1,2},\ldots,X_{1,n}$ are the items in iteration 1, then $X_{2,1}, X_{2,2},\ldots,X_{2,n}$ are the items on iteration 2, and so on. The overall sum would then be written $\sum_{k=1}^N\sum_{i=1}^n f(X_{k,i})$.
Best Answer
It has the following meaning $$\sum_{i,j=1}^n=\sum_{i=1}^n\sum_{j=1}^n=\sum_{j=1}^n\sum_{i=1}^n$$ Since the sums are finite, the order of summation does not matter. This notation is convenient because it takes up less space than writing it out.