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How one can prove $$\mathbb{E}[W_{t}^{2k}]=\frac{(2k)!}{2^kk!}t^k,$$

where $\{W_t\}_{t\ge 0}$ is a standard Brownian motion on some filtered probability space $(\Omega , \mathcal{F}_{t}, \{\mathcal{F}_{t}\}_{t\ge 0}, \mathbb{P})$ and $k$ is a positive integer?

I know that it is possible by using induction and Ito formula. Here I want to prove based on standard results. I mean using normally distributed function and moment function. Thanks!

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    For a fixed $t$, $W_t$ is just a normal random variable with mean zero and variance $t$. Do you know how to find the moments of a normal random variable? – 2017-02-10
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    @carmichael561 Moment formula :$M^{(k)}(0)=E(X^k)=k!f(k)$. But, I couldn't go further – 2017-02-10
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    I think we get use Ito's formula to get that isn't it? but I did see the above lint has used that formula. @ armichael561 – 2017-05-08

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For fixed $t>0$, $W_t$ is a normal random variable with mean $0$ and variance $t$, and has pdf $$ f(x)=\frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}} $$ Using this pdf and completing the square, one can show that the moment generating function for $W_t$ is $$ M(s)=\mathbb{E}[e^{sW_t}]=\exp\Big(\frac{ts^2}{2}\Big) $$ On the other hand, $$ M(s)=\sum_{k=0}^{\infty}\frac{\mathbb{E}[W_t^k]}{k!}s^k $$ so by comparing coefficients it follows that $\mathbb{E}[W_t^k]=0$ if $k$ is odd, while $$ \mathbb{E}[W_t^{2k}]=\frac{(2k)!}{2^kk!}t^k $$