The pattern actually is
$$ \sum_{n_m=1}^{n}\sum_{n_{m-1}=1}^{n_m}\ldots
\sum_{n_1=1}^{n_2} \sum_{n_0=1}^{n_1} 1
= \frac{1}{(m+1)!}\prod_{k = 0}^{m}(n+k), \tag1$$
where for reasons of symmetry (and making the later proof simpler)
I have written $n_1$ as $\sum_{n_0=1}^{n_1} 1.$
A slightly more convenient way to write the same thing is
$$ \sum_{1\leq n_0\leq n_1\leq n_2\leq \cdots \leq n_{m-1}\leq n_m\leq n} 1
= \binom{n+m}{m+1} \tag2$$
where $\binom{n+m}{m+1}$ is a binomial coefficient.
The right-hand side of Equation $2$ equals the right-hand side of Equation $1$ by means of the following formula for a binomial coefficient,
$$
\binom pq = \frac{p(p-1)(p-2)\cdots(p-q+1)}{q!}.
$$
The meaning of the left-hand side of Equation $2$ is that there is one term of the sum for every possible list of numbers $n_0, n_1, n_2, \ldots, n_m$ such that $1\leq n_0\leq n_1\leq n_2\leq \cdots \leq n_m\leq n.$
Notice that
$$ \sum_{1\leq n_0\leq n_1\leq n_2\leq\cdots\leq n_{m-1}\leq n_m\leq n} 1
= \sum_{n_m=1}^n \left(\sum_{1\leq n_0\leq n_1\leq n_2\leq\cdots
\leq n_{m-1}\leq n_m} 1\right),$$
and if you continue to "unpack" the sums in this fashion with a sum from $1$ to $n_m,$ then $1$ to $n_{m-1},$ and so forth, you get the $m+1$
nested sums on the left side of Equation $1.$
This is a well-known result.
See Simplification of a nested sum,
Nested summations and their relation to binomial coefficients,
and this answer to
Binomial coefficient as a summation series proof?
There is a combinatorial proof which is a little easier to see if you
rewrite the sum this way:
$$ \sum_{1\leq n_0\leq n_1\leq n_2\leq\cdots\leq n_m\leq n} 1
= \sum_{0 < n_0 < n_1+1 < n_2+2 < \cdots < n_m+m < n+m+1} 1,\tag3$$
using the fact that for integers $p$ and $q,$ $p \leq q$ if and only if
$p < q+1.$
Each term in the sum on the right-hand side of Equation $3$
has $m+1$ index numbers $n_0, n_1, n_2, \ldots, n_m$
selected from the integers strictly between $0$ and $n+m+1,$ that is,
from the set of integers $\{1,2,3,\ldots,n+m-1,n+m\}.$
Since each possible combination of numbers selected can be selected in only
one way (increasing order), the number of terms is exactly the number
of ways to choose $m+1$ elements from a set of $n+m$ elements, that is,
the binomial coefficient "$n+m$ choose $m+1$," notated
$\binom{n+m}{m+1}.$