Is $n^{1/n}$ irrational for all integers $n \geq 2$, and does $\epsilon_n = \frac{1}{n}(1-n^{-1/n})$ have known number-theoretic properties?

Background: I was exploring whether the standard normalization $\sum p_i = 1$ in probability is exact when the parts are written as real numbers. This led me to the quantity: $$\epsilon_n = \frac{1}{n}\left(1 - n^{-1/n}\right)$$ which represents a "gap" when dividing $1$ into $n$ equal parts.

What I know: I attempted to prove that $\epsilon_n$ is always irrational for $n \geq 2$:

Step 1: Show $n^{1/n}$ is irrational for $n \geq 2$. Assume $n^{1/n} = \frac{p}{q}$ with $p,q \in \mathbb{Z}^+$ and $\gcd(p,q)=1$. Then $n = \frac{p^n}{q^n}$, so $nq^n = p^n$.

• If $q \geq 2$: then $q \mid p^n$, which implies $q \mid p$, contradicting $\gcd(p,q)=1$.
• If $q = 1$: then $n = p^n$. Since we are looking for integers $n \geq 2$, if $p \geq 2$ we get $p^n \geq 2^n > n$. Thus, there are no integer solutions.

Therefore, $n^{1/n} \notin \mathbb{Q}$ for $n \geq 2$.

Step 2: Show $\epsilon_n$ is irrational. We have $\epsilon_n = \frac{1}{n} - \frac{1}{n}(n^{-1/n})$. Since $n^{1/n} \notin \mathbb{Q}$, its reciprocal $n^{-1/n} \notin \mathbb{Q}$. Multiplying this by the non-zero rational $\frac{1}{n}$ yields an irrational number. Subtracting this irrational from the rational number $\frac{1}{n}$ guarantees that $\epsilon_n \notin \mathbb{Q}$.

My questions:

1. Is this proof correct?
2. Does $\epsilon_n$ appear anywhere in number theory or have a specific name?
3. Is there a deeper structure behind $\epsilon_n \in \mathbb{R} \setminus \mathbb{Q}$ for all $n \geq 2$ in the context of probability or measure theory?