Prove that $\varphi(2n)=2\varphi(n)$ if $n$ is even. So, far, I have the following which I have seen the basic idea in other similar proofs:

Let $n$ be an even, positive integer. Then $n=2^{s}(p_{1}^{x_{1}}\cdot p_{2}^{x_{2}} \cdot\cdot\cdot p_{r}^{x_{r}})$ since $n$ is even, where $p_{r}$ is an odd prime number and $x$ is an positive integer. Then $\frac{n}{2^{s}}$ is odd, and so it is relatively prime to $2^{s}$ which is even. But $n=2^{s}\cdot \frac{n}{2^{s}}$, which implies that $\varphi(n)=\varphi(2^{s}) \cdot \varphi(\frac{n}{2^{s}})$.

I am hoping to somehow get $2$ out of $\varphi(2^{s})$ and combine the leftovers with the term $\varphi(\frac{n}{2^{s}})$, but not sure if it's possible? It seems like I keep end up proving the statement false, when it should be true...is there a clear way to finish this, or is there an easier way to do it?
Write $n = 2^k\cdot m$ where $m$ is odd. Then
$$\varphi(2n) = \varphi(2^{k+1} \cdot m) = \varphi(2^{k+1})\varphi(m) = 2^k\varphi(m) = 2\varphi(2^k)\varphi(m) = 2\varphi(n).$$
Since $n$ is odd, $\gcd(2,n)=1$. 


  $$\varphi(2n) = \varphi(2)\varphi(n) = 1\varphi(n)$$ 


Since $n$ is even, $n = 2^km$


  $$\varphi(2n) = \varphi(2\cdot 2^km) = \varphi(2^{k+1}m) = \varphi(2^{k+1})\varphi(m)= 2^k\varphi(m)= $$$$2\cdot 2^{k-1}\varphi(m)=2\cdot \varphi(2^k)\varphi(m)=2\cdot \varphi(2^km) = 2\varphi(n)$$