Diffie-Hellman Shared Secret Agreement

By definition $A = g^a \bmod p$ and $B = g^b \bmod p$ are the results of modular exponentiation, so Bob receives $A$ and computes $B^a = (g^b)^a \bmod p$, while Alice receives $B$ and computes $A^b = (g^a)^b \bmod p$.

Modular exponentiation obeys the same law of exponents as ordinary exponentiation, because repeated multiplication modulo $p$ composes exactly like repeated multiplication of integers, only reduced modulo $p$ at each step: $(g^b)^a = g^{ba} = g^{ab} = (g^a)^b$ as exponents of $g$, and this equality survives reduction modulo $p$ at every stage.

Therefore $B^a \bmod p = g^{ab} \bmod p = A^b \bmod p$: Alice and Bob independently compute the same quantity $g^{ab} \bmod p$ from their own secret exponent and the other party's public value, without either secret $a$ or $b$ ever appearing on the channel.

The security argument is separate from this correctness argument: correctness only shows both parties land on the same number; hardness of computing $a$ from $g$, $p$, and $A$ (the discrete logarithm problem) is what keeps that shared number secret from an observer.