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

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.

From finite calculus we have that

$$\sum k^{\underline m}\delta k=\frac{k^{\underline{m+1}}}{k+1}+C$$

where $a^{\underline k}:=\prod _{j=0}^{k-1}(a-j)$ is known as a falling factorial, and $C$ is any periodic function with period $1$ (this can be a constant function, in general is taken as zero, this is an analog of an indefinite integral, in this case this is an indefinite sum).

And we have that $a^{\overline m}:=\prod_{j=0}^{m-1}(a+j)$ is known as a rising factorial, and

$$a^{\overline m}=(a+m-1)^\underline m$$

Hence you want to solve the sum

$$\begin{align}\sum_{k=\ell}^n k(k+1)\cdots(k+m)&=\sum\nolimits_\ell^{n+1}k^{\overline{m+1}}\delta k\\&=\sum\nolimits_\ell^{n+1}(k+m)^{\underline{m+1}}\delta k\\&=\frac{(k+m)^{\underline{m+2}}}{m+2}\bigg|_\ell^{n+1}\\&=\frac1{m+2}\big((n+m+1)^\underline{m+2}-(\ell+m)^\underline{m+2}\big)\\&=\frac1{m+2}\big(n^\overline{m+2}-(\ell-1)^\overline{m+2}\big)\end{align}$$

From here is easy to justify your result

$$\underbrace{\sum\sum\ldots\sum_{k=1}^n 1}_{m\text{ times}}=\frac{n^\overline {m+1}}{m!}=\frac{(n+m-1)^{\underline m}}{m!}=\binom{n+m-1}{m}$$