This must be covered in almost every text on the $p$-adic numbers; I think the book of Gouvêa is the best of these.

And your statement is not quite true: for $p=2$, $\mu_{p-1}$ is trivial all right, but the other part is isomorphic to $\{\pm1\}\times(1+4\Bbb Z_2)$. My tale below omits the story for $p=2$, and you can fill this in yourself.

First, you can consider the units, $\Bbb Z_p^\times$, and reduce them modulo $p$ to the multiplicative group of $\Bbb F_p\cong\Bbb Z/p\Bbb Z$. It’s cyclic of order $p-1$, as I’m sure you know. So we have an exact sequence:
$$
0\longrightarrow K\longrightarrow\Bbb Z_p^\times\longrightarrow\Bbb F_p^\times\longrightarrow0\,;
$$
if you’re unfamiliar with the notation of exact sequences this merely says that there’s a surjective map from the middle term to the one to its right, with kernel equal to the one to its left.

What’s the kernel? It’s the units that go to $1$ in the field $\Bbb F_p$, in other words $1+p\Bbb Z_p$. To prove that $\Bbb Z_p^\times$ is the direct product of the two things to either side of it, it’s enough to show that there’s a homomorphism from $\Bbb F_p^\times$ into it whose image hits $K$ only in the identity. This is the fun part:

You can find $(p-1)$-th roots of unity in $\Bbb Z_p$ either by a routine application of any version of Hensel’s Lemma that you like, or, my favorite method, take an element of $\Bbb F_p^\times$, lift it to *any* element of $\Bbb Z_p$ that goes to it modulo $p$, and take successive $p$-th powers: $x\mapsto x^p\mapsto(x^p)^p\mapsto\cdots$ etc. I’ll leave it to you to show that this is a good convergent sequence, and its limit is clearly a suitable root of unity.

Showing that the multiplicative group $1+p\Bbb Z_p$ is isomorphic to the additive group $\Bbb Z_p^+$ is rather less fun. To tell you the truth, when I first saw the logarithmic argument, I didn’t like it, but I now think it’s the best one. You have to convince yourself that as long as the series for $\log(1+x)$ that you saw in Calculus is convergent, then $\log\bigl[(1+x)(1+y)\bigr]=\log(1+x)+\log(1+y)$, and for $x\in p\Bbb Z_p$, the series is convergent. And the values of the log are all in $p\Bbb Z_p\cong\Bbb Z_p$ (as additive groups, of course), and fill out that group.