The multiplicity of a root $r$ of a irreducible polynomial is a power of $p$ characteristic $f$ is an irreducible polynomial over a field $K$ of characteristic $p$. $F$ is a splitting field of $f$ over $K$ and $u_1$ a root of $f$.
I have shown that $f=[(x-u_1)\cdots (x-u_n)]^{[K(u_1):K]_s}$ and $n=[K(u_1):K]_i$, where $[K(u_1):K]_s$ is separable degree and $[K(u_1):K]_i$ is inseparable degree.
Then I have trouble showing that $u^{[K(u_1): K]_i}$  is separable.
One proof points out that $[K(u_1):K]_i = p^k = r$ whence $f=[(x-u_1)\cdots (x-u_n)]^{[K(u_1):K]_s} = (x^r - u_1^r) \cdots (x^r - u_n^r)$ is of $K[x]$ therefore $(x-u_1^r) \cdots (x-u_n^r)$ is of $K[x]$ with $u_1^r$ $\cdots$ $u_n^r$ distinct so $u_1^r$ is irreducible.
Then I have a problem. Why is $[K(u_1):K]_i$ is a power of $p$.
 A: At the beggining you mixed separable and inseparable degree. The main fact here is: If $K$ is of characteristic $p$, and $f\in K[x]$ is irreducible, then there exist $k\geqslant 0$ and an irreducible and separable polynomial $f_s\in K[x]$ such that $f(x)=f_s(x^{p^k})$. If $\alpha$ is a root of $f$, then the separable degree of $K(\alpha)/K$ is $[K(\alpha):K]_s=\deg(f_s)$ and the inseparable degree of $K(\alpha)/K$ is $[K(\alpha):K]_i=p^k$. Note that $[K(\alpha):K]= \deg(f)= p^k\deg(f_s)$, so it doesn't have to be the power of $p$.
Also, $\beta= \alpha^{p^k}$ is separable because its minimal polynomial is $f_s(x)$: $f_s(\beta)= f_s(\alpha^{p^k})= f(\alpha)=0$ and $f_s(x)$ is irreducible and separable.
Maybe you can look at the following example: Take $p\neq 2$ and $K=\mathbb F_p(t)$, where $t$ is transendental over $\mathbb F_p$. Consider polynomial $f(x)= x^{2p}-t$. It is irreducible over $K$, and for it we have $f_s(x)=x^2-t$ and $k=1$, and $f(x)= f_s(x^p)$. ($f_s(x)$ is separable because $p\neq 2$.) So, if $\alpha$ is one of the roots of $f(x)$, then $[K(\alpha):K]=2p$, $[K(\alpha):K]_s=2$ and $[K(\alpha):K]_i=p$.
You can investigate this situation more explicitly. Since $\alpha^{2p}=t$, we can write $f(x)= x^{2p}-\alpha^{2p}= (x^p-\alpha^p)(x^p+\alpha^p)= (x-\alpha)^p(x+\alpha)^p$. So $f(x)$ has two distinct roots: $\alpha$ and $-\alpha$ (thus separable degree is $2$), both of which are of multiplicity $p$ (thus inseparable degree is $p$). 
