Proving function is $C^k$ This question is from an exercise in Way of Analysis (section 10.2.4 problem 20).
If $f: \mathbb{R} \rightarrow \mathbb{R}$ is $C^k$ and $f$ is even, then show $F: \mathbb{R}^n \rightarrow \mathbb{R}$ given by $F(x) = f(|x|)$ is also $C^k$.
$C^k$ just means the function has $k$ continuous derivatives, and $|x|$ is referring to the regular Euclidean norm.
I am not sure how to use the fact $f$ is even in the proof and having trouble for a general $k$. Any hints, even on a particular case like $k=2$, are welcome.
My workings:
As a comment suggests $f(x) = F(x)$ for $n = 1$. For $n = 2$, $F(x) = f(\sqrt {x_1^2 + x_2^2}).$ I'm stuck on where to proceed from here. Is there some form of induction here that I am not seeing?

Here's a start: Suppose $f \in C^2(\mathbb {R}).$ Note $f$ even implies $f'$ is odd. This is key, because it tells us$f'(0)=0.$ So letting $P_2$ denote the second degree Taylor polynomial of $f$ at $0,$ we have $P_2(t) = f(0) + f''(0)t^2/2.$ Clearly $P_2(|x|)\in C^\infty(\mathbb {R}^n).$
Set $g(t)= f(t)-P_2(t).$ Then $g\in C^2$ and $0=g(0)=g'(0)=g''(0).$ So it's enough to check that $G(x) = g(|x|)$ is $C^2.$ We certainly have $G \in C^2(\mathbb {R}^n\setminus \{0\}).$ Suppose $D$ is a partial derivative of order $\le 2.$ If we show that $D(G)$ tends to $0$ at the origin, then it will follow that $G \in C^2(\mathbb {R}^n).$
