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.

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

The summation:

Becomes the numerator in the following fraction:

And the fraction simplifies to the most common form of the answer:

Now I can sleep.

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.

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

The summation:

Becomes the numerator in the following fraction:

And the fraction simplifies to the most common form of the answer:

Now I can sleep.