Prove $p_0-p_2+p_4-\cdots=2^{n/2}\cos{\dfrac{n\pi}{4}}$ and $p_1-p_3+p_5-\dots=2^{n/2}\sin{\dfrac{n\pi}{4}}$

Consider: $$(1+x)^n= p_0 + p_1 x + p_2 x^2+\cdots$$ From where $$p_0=1,\quad p_1=\dfrac{n}{1},\quad p_2=\dfrac{n(n-1)}{2!},\ldots$$ Are the coefficients of the Newton´s Binomial expansion, using $x=i$ show that:

$$p_{0}-p_{2}+p_{4}-\dots=2^{n/2}\cos{\dfrac{n\pi}{4}}$$

$$p_{1}-p_{3}+p_{5}-\dots=2^{n/2}\sin{\dfrac{n\pi}{4}}$$

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Adding to Kuai's answer, Use De Moivre's theorem to expand on $(1+i)^n$
$$2^{n/2}*cos(\frac{n\pi}{4})$$ as the real part and
$$2^{n/2}*sin(\frac{n\pi}{4})$$ as the imaginary part
Hint: $$1+i=2^{1/2}(\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4}))$$ Also, the two sums of interest are the real and imaginary parts of $(1+i)^n$, respectively.
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