Let $K$ be a Sylow subgroup of a finite group $G$. Prove that if $x \in N(K)$ and the order of $x$ is a power of $p$, then $ x \in K$. Let $K$ be a Sylow subgroup of a finite group $G$. Prove that if $x \in N(K)$ and the order of $x$ is a power of $p$, then $ x \in K$. 

This is how I tried...

Since $K$ is normal in $N(K)$, the factor group $$\frac{N(K)}{K}=\{K,xK,....\}
$$ is justified. 

Then I took $$(xK)^{P^t}=x^{P^t}K=K$$. 

This is where I am stuck.
You can also do this in a more elementary way not using the quotient group.

As already noted $K$ is normal in $N(K)$. This implies that $K$ is the only $p$-Sylow-subgroup of $N(K)$ (since $N(K)$ certainly has no $p$-subgroups of larger order and all $p$-Sylow-subgroups of $N(K)$ are conjugate under $N(K)$). Hence the group generated by $x$ (a $p$-subgroup of $N(K)$) lies in $K$ (since it must lie in SOME $p$-Sylow-subgroup of $N(K)$ and there is only one).
Clearly, $K$ is normal in $N(K)$ so we can talk about the factor group $\frac{N(K)}{K}$. Since $G$ is finite and K is a Sylow p-subgroup of $G$ so $\frac{|N(K)|}{|K|}$ doesn't have any power of $p$ and it is given that $|x|$ is a power of $p$ so $\gcd(|x|,\frac{|N(K)|}{|K|}) = 1$ $\implies \exists \ m,n\in \mathbb{Z}$ such that $|x|m + n\frac{|N(K)|}{|K|} = 1$. and since  $x\in N(K)$ we know that $x^{\frac{|N(K)|}{|K|}} \in K$. Now $x = x^{|x|m + n\frac{|N(K)|}{|K|}} = (x^{|x|})^m (x^{\frac{|N(K)|}{|K|}})^n = (x^{\frac{|N(K)|}{|K|}})^n \in K$ as $x^{\frac{|N(K)|}{|K|}} \in K$