Strictly Increasing Integers

An easier approach is to see there is one such integer for each nonempty subset of $\{1,2,3,4,5,6,7,8,9\}$. Once you have a subset, you arrange the digits in increasing order, so there are $2^9-1$ integers with strictly increasing digits.

Lets say the number has $k$ digits, all distinct. We notice, that out of all possible permutations of the digits, only one satisfies the condition of the problem, that is, strictly ascending.

If we would try to generate a solution ourselves, we would pick digits from 1 to 9 $k$ times, except we would not be allowed to pick the same digit twice. The number of ways to do this for a fixed $k$ is the number of combinations

$C(n, k) = \frac{n!}{k!(n-k)!}$

where $n=9$ is the total number of digits we choose from

The total number of solutions is sum over number of solutions for a fixed digit length

$S = \sum_{k=1}^{n} C(n, k)$

As mentioned here

https://en.wikipedia.org/wiki/Binomial_coefficient#Sum_of_coefficients_row

There is a theorem for summing up binomial coefficients $C(n,k)$, which states that

$S = 2^n - 1 = 511$ in your case