Ito-Doeblin Formula Question

[BWV]

Reading through a proof on why the higher order terms vanish and it makes this statement

dW(t)dW(t) = dt

where W(t) is a Brownian motion

It is not obvious to me why this is the case, but the text seems to infer that it is because no further explanation is offered

[Focus]

It may have to do with W_t=\sqrt{t} B where B is N(0,1).

[gel]

\delta W_t \sim N(0,t). It follows that E[(\delta W_t)^2]=\delta t and E[|\delta W_t|^3]={\rm const}\times \delta t^{3/2}. So third and higher powers of dW are smaller order than dt on average , and therefore vanish if you sum them over a partition and let dt->0.