I didn't need to attempt to work out the integral version of the summation I'd found that represented the number of times at least one 1 would appear when rolling an n-sided die n times.

Instead, there is a direct answer, right in one of the tables I'd shown, for what my summation becomes.

Specifically, I noticed that there was a pattern to the differences between the numerator and denominator values that represented the odds for each listed n. Following up on it this time, I noticed that the difference was (n-1)^n.

Thus, the numerator would be the denominator minus the difference, aka:

n^n - (n-1)^n

Which, when divided by the denominator (n^n) yields the familiar formula for the situation at hand.

![showing the numerator and denominator differences](https://raw.githubusercontent.com/slowerthanlightspeed/reddit_images/master/tetrations%20connection%20solved.png)

(side note, Wolfram Alpha verified for me that the summation becomes n^n - (n-1)^n)

The summation:
![The Summation](https://raw.githubusercontent.com/slowerthanlightspeed/slowerthanlightspeed.github.io/master/reddit_images/sum_for_probability.png)

Becomes the numerator in the following fraction:
![Odds Fraction](https://github.com/slowerthanlightspeed/reddit_images/blob/master/last%20major%20step%20between%20quick%20solution%20from%20tetrations.png?raw=true)

And the fraction simplifies to the most common form of the answer:
![common solution format](https://github.com/slowerthanlightspeed/reddit_images/blob/master/most_common_form_solution.png?raw=true)

Now I can sleep.