Wiener Process $dB^2=dt$ Why is $dB^2=dt$? Every online source I've come across lists this as an exercise or just states it, but why isn't this ever explicitly proved? I know that $dB=\sqrt{dt}Z$, but I don't know what squaring a Gaussian random variable means.
$$dB_t^2 = dt, \qquad (dt)^2 = 0, \qquad dB_t \, dt = 0 \tag{1}$$ are basically rules to simplify the calculation of the quadratic (co)variation of Itô processes - and nothing more:

Let $(B_t)_{t \geq 0}$ a one-dimensional Brownian motion and $(X_t)_{t \geq 0}$ an Itô process, i.e.

$$dX_s = \sigma(s) \, dB_s + b(s) \, ds$$

Then, by Itô's formula,

$$f(X_t)-f(X_0) = \int_0^t f'(X_s) \, dX_s + \int_0^t f''(X_s) \, \sigma^2(s) \, ds. \tag{2}$$

The point is: If we simply apply the rules in $(1)$, we obtain

$$dX_s^2 = (\sigma(s) \, dB_s + b(s) \, ds)^2 = \sigma^2(s) \underbrace{dB_s^2}_{ds} + 2b(s) \sigma(s) \underbrace{ds B_s}_{0} + b^2(s) \, \underbrace{ds^2}_{0} \\ = \sigma^2(s) \, ds.$$

Therefore, we can rewrite $(2)$ in the following way:

$$f(X_t)-f(X_0) = \int_0^t f'(X_s) \, dX_s + \int_0^t f''(X_s) \, dX_s^2$$

i.e. Itô's formula justifies the calculation rules in $(1)$.



The (mathematical) reason why this works fine can be seen while proving Itô's formula. Actually, one can show that

$$\sum_{j=1}^n g(B_{t_{j-1}}) \cdot (B_{t_j}-B_{t_{j-1}})^2 \tag{3}$$

converges to

$$\int_0^t g(B_s) \, ds$$

as the mesh size $|\Pi|$ of the partition $\Pi=\{0=t_0&lt;\ldots&lt;t_n=t\}$ tends to $0$. This convergence is based on the fact that $B_t^2-t$ is a martingale. On the other hand, comparing $(3)$ with the definition of Riemann-Stieltjes integrals, it's natural to define

$$\int_0^t g(B_s) \, dB_s^2 := \lim_{|\Pi| \to 0}\sum_{j=1}^n g(B_{t_{j-1}}) \cdot (B_{t_j}-B_{t_{j-1}})^2.$$

Consequently,

$$\int_0^t g(B_s) \, dB_s^2 = \int_0^t g(B_s) \, ds.$$

A similar reasoning applies to Itô processes. Note that these integrals, i.e. integrals of the form

$$\int_0^t g(X_s) \, dX_s^2,$$

are exactly the integrals appearing in Itô's formula $(2)$ ($g \hat{=} f''$).
Obviously $dB_t^2 \neq dt$, since $dB_t \sim \mathcal{N} (0, dt)$ is a random variable, while $dt$ is deterministic.

As Michael Hardy said, they really meant to say $\mathbb{E} \left[ dB_t^2 \right] = dt$.  To convince yourself, compute 
$$ \mathbb{E} \left[ dB_t^n \right] = \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2 \pi dt}} \exp\left(-\frac{x^2}{2  dt}\right) x^n dx \, .$$