$(\mathbb{Z}/p^n \mathbb{Z})^{\times}$ is the group of invertible elements of $\mathbb{Z}/p^n \mathbb{Z}$.
To find the order of $1+p$, we start by expanding $(1 + p)^{p}$ using the binomial theorem:
$(1 + p)^p = 1 + p \cdot p + {p \choose 2}\cdot p^2 + \dots \equiv 1 + p^2 \pmod {p^3}$. Similarly,
$(1 + p)^{p^2} \equiv 1 + p^3 \pmod {p^4}$
$\cdots$
$(1 + p)^{p^{n-1}} \equiv 1 + p^{n} \pmod{p^{n+1}}$.
I'm thrown off by $\pmod{p^{n+1}}$. I need $p^{n}$ where there is $p^{n+1}$.
If you accept those identities, then $(1+p)^{p^{n-1}} \equiv 1+p^n \pmod{p^{n+1}}$ implies $(1+p)^{p^{n-1}} \equiv 1 \pmod{p^n}$, so the order divides $p^{n-1}$, so the order is a power of $p$, but the lower powers of $p$ cannot give $1$ by the lower identities.