calculus - Is there a pattern to expression for the nested sums of the first $n$ terms of an expression?

Apologies for the confusing title but I couldn't think of a better way to phrase it. What I'm talking about is this:

$$ \sum_{i=1}^n \;i = \frac{1}{2}n \left(n+1\right)$$
$$ \sum_{i=1}^n \; \frac{1}{2}i\left(i+1\right) = \frac{1}{6}n\left(n+1\right)\left(n+2\right) $$
$$ \sum_{i=1}^n \; \frac{1}{6}i\left(i+1\right)\left(i+2\right) = \frac{1}{24}n\left(n+1\right)\left(n+2\right)\left(n+3\right) $$

We see that this seems to indicate:

$$ \sum_{n_m=1}^{n}\sum_{n_{m-1}=1}^{n_m}\ldots \sum_{n_1=1}^{n_2} \; n_1 = \frac{1}{m!}\prod_{k = 0}^{m}(n+k) $$

Is this a known result? If so how would you go about proving it? I have tried a few inductive arguments but because I couldn't express the intermediate expressions nicely, I didn't really get anywhere.

Answer

You should have
$$\sum_{i=1}^{n} 1 = n$$
$$\sum_{i=1}^{n} i = \frac{1}{2} n(n+1)$$
$$\sum_{i=1}^{n} \frac{1}{2} i(i+1) = \frac{1}{6} n(n+1)(n+2)$$
$$\sum_{i=1}^{n} \frac{1}{6} i(i+1)(i+2) = \frac{1}{24} n(n+1)(n+2)(n+3)$$

In particular, the first sum of yours was wrong and the things you were adding should depend on $i$, not on $n$.

But, to answer the question, yes! This is a known result, and actually follows quite nicely from properties of Pascal's triangle. Look at the first few diagonals of the triangle and see how they match up to your sums, and see if you can explain why there's such a relation, and why the sums here can be written in terms of binomial coefficients. Then, the hockey-stick identity proves your idea nicely.