Induction Proof of Taylor Series Formula I'm attempting to prove a formula for the taylor series of function from a differential equation. 
The equation is
$$f(0)=1$$ $$f'(x) = 2xf(x)$$
I have found empirically that
$$f(x) = \sum_{k=0}^{\infty}\frac{x^{2k}}{k!}$$
 I need to prove that this general formula works via induction.
Here is my attempt!
$$\mathrm{Show} \qquad1+x^2+\frac{x^4}{2}+\frac{x^6}{6}+...+\frac{x^{2k}}{k!}=\sum_{n=0}^k \frac{f^{(2n)}(0)x^{2n}}{(2n)!}$$

Prove true for $k=0$

$$1=\frac{f^{(0)}(0)x^{0}}{(0)!}$$
$$1 = 1 \ \checkmark$$

Assume true for $k=c$

$$1+x^2+\frac{x^4}{2}+...+\frac{x^{2c}}{c!}=\sum_{n=0}^c \frac{f^{(2n)}(0)x^{2n}}{(2n)!}$$

Prove true for $k=c+1$

$$\begin{align} 1+x^2+\frac{x^4}{2}+...+\frac{x^{2c}}{c!}+\frac{x^{2c+2}}{(c+1)!}&=\sum_{n=0}^{c+1} \frac{f^{(2n)}(0)x^{2n}}{(2n)!} \\
\sum_{n=0}^c \frac{f^{(2n)}(0)x^{2n}}{(2n)!}+\frac{x^{2c+2}}{(c+1)!}&=\sum_{n=0}^{c+1} \frac{f^{(2n)}(0)x^{2n}}{(2n)!} \\ \frac{x^{2c+2}}{(c+1)!}&=\frac{f^{(2c+2)}(0)x^{2c+2}}{(2c+2)!}
\end{align}$$
From there I don't know how to proceed. Maybe I shouldn't have broken apart the sum on the right? Any help would be appreciated. 
Note: I know the differential is easily separable and solvable, but the project involves comparing the solutions
 A: Note: This proof relies on a formula for the $n^{th}$ derivative of $f(x)$ that can be found HERE
The proof is rather extensive, and would roughly quadruple the length of this answer.
Proof  of The Taylor Expansion of $f(x)$
$$\mbox{Prove true that } \ 1 + x^2 + \frac{x^4}{4} + \frac{x^6}{6} + \cdots + \frac{x^{2k}}{k!}=\sum_{n=0}^{k}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!}$$
By Induction: 
$\circ \text{ Prove true for }k=0$
$$\frac{f^{(0)}(0)x^{0}}{0!} \\ 1=1 \ \checkmark$$
$\circ \text{ Assume true for }k=c$
$$ 1 + x^2 + \frac{x^4}{4} + \frac{x^6}{6} + \cdots + \frac{x^{2c}}{c!}=\sum_{n=0}^{c}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!}$$
$\circ \text{ Prove true for }k=c+1$
\begin{equation} 
\begin{split}
 1 + x^2 + \frac{x^4}{4} + \frac{x^6}{6} + \cdots + \frac{x^{2c}}{c!}+\frac{x^{2c+2}}{(c+1)!}&=\sum_{n=0}^{c+1}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!} \\ \sum_{n=0}^{c}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!}+\frac{x^{2c+2}}{(c+1)!}&=\sum_{n=0}^{c+1}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!}  \mbox{By I.H}
 \\ =\sum_{n=0}^{c}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!}+\frac{x^{2c+2}}{(c+1)!} \cdot \frac{(2c+2)!}{(2c+2)!}
 \\=\sum_{n=0}^{c}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!}+\frac{x^{2c+2}}{(2c+2)!} \cdot \frac{(2c+2)!}{(c+1)!}
 \\=\sum_{n=0}^{c}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!}+\frac{x^{2c+2}}{(2c+2)!} \cdot \frac{(2c+2)!}{(\frac{2c+2}{2})!}
 \\=\sum_{n=0}^{c}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!}+\frac{x^{2c+2}}{(2c+2)!} \cdot f^{(2c+2)}(0) \mbox{ From linked proof}
 \\=\sum_{n=0}^{c+1}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!} \checkmark
\end{split}
\end{equation} 
$$f(x)=\sum_{k=0}^{\infty} \frac{x^{2k}}{k!} \mathrm{Q.E.D}$$
A: Once you have your formula,
the proof is straightforward.
If
$f(x) 
= \sum_{k=0}^{\infty}\frac{x^{2k}}{k!}
$,
then
$\begin{array}\\
f'(x) 
&= \sum_{k=0}^{\infty}\frac{2kx^{2k-1}}{k!}\\
&= \sum_{k=1}^{\infty}\frac{2kx^{2k-1}}{k!}
\qquad\text{since the }k=0\text{ term is zero}\\
&= \sum_{k=1}^{\infty}\frac{2x^{2k-1}}{(k-1)!}
\qquad\text{cancelling }k\\
&= \sum_{k=0}^{\infty}\frac{2x^{2k+1}}{k!}
\qquad\text{shifting }k\\
&= \sum_{k=0}^{\infty}\frac{2xx^{2k}}{k!}\\
&= 2x\sum_{k=0}^{\infty}\frac{x^{2k}}{k!}\\
&=2xf(x)\\
\end{array}
$,
