Simplicity of the roots of a minimal polynomial Let $L/K$ be a finite field extension, and let $\mu_{\alpha,K}\in K[X]$ be the minimal polynomial of $\alpha\in L$.
One can easily see that $\alpha$ is a simple root of $\mu_{\alpha,K}$. Indeed, if it weren't the case, the polynomial $\mu_{\alpha,K}'\in K[X]$ would vanish at $\alpha$ but has lower degree that $\mu_{\alpha,K}$, which contradicts minimality of $\mu_{\alpha,K}$.
To go further, how would one prove that all roots $\beta$ of $\mu_{\alpha,K}$ are simple ?
 A: liaombro has ably answered the question; I write to add that you have a hidden assumption in your argument: that the characteristic of $K$ does not divide $\deg(\mu_{\alpha,K})$. 
If $K$ is a field and $L$ is an extension, then an algebraic element $\alpha\in L$ is separable over $K$ if the minimal polynomial $\mu_{\alpha,K}$ does not have repeated roots. In characteristic zero, all algebraic elements are always separable. However, in positive characteristic, there are algebraic elements that are not separable. However, if $K$ is a finite field, then $\alpha$ is always separable, so you may not have seen examples where the situation occurs.
Your argument breaks down if $\mu_{\alpha,K}’ = 0$. Since the minimal polynomial is monic, this will occur only if the characteristic of $K$ divides the degree of $\mu_{\alpha,K}$. In fact, one can show that if $\mathrm{char}(K)=p\gt 0$, then algebraic element $\alpha$ is inseparable (not separable) if and only if the minimal polynomial $\mu_{\alpha,K}$ is of the form $g(x^p)$ for some polynomial $g(x)$; that is, it’s a polynomial in $x^p$.
For an explicit example, let $K=\mathbb{F}_p(x)$, the field of rational functions with coefficients in the field of $p$ elements. Let $g(Y) = Y^p-x\in K[Y]$. Let $L$ be the extension obtained by adjoining a root of $g(Y)$ to $K$; call this root $\alpha$ (that is, $\alpha$ is a $p$th root of $x$). It is not hard to verify that $g(Y)$ is irreducible (e.g., use Eisenstein’s Criterion in $\mathbb{F}_P[x][Y]$), so that $g(Y)$ is the monic irreducible polynomial of $\alpha$ over $K$. However, in $L$ we have $g(Y) = Y^p-x =(Y-\alpha)^p$, so that $g(Y)$ has a single root, repeated $p$ times. You can see that $g’(Y)=0$, so that’s where your argument breaks down. 
A: Take the minimal polynomial of $\beta$, that is, $\mu_{\beta, K}$ and consider $gcd(\mu_{\alpha, K}, \mu_{\beta, K})$. That gcd has $\beta$ as a root, so it must be $\mu_{\beta, K}$. Since $\mu_{\alpha, K}$ is irreducible, $\mu_{\alpha, K}=\mu_{\beta, K}$ and we're done.
