Proof of the Diffie-Hellman Key Exchange

Let $g^a = x bmod p$ then $k in mathbb{Z}$; $$g^a = x + k cdot p$$ Now take $b$-th power of both sides and $bmod p$.

begin{align} (g^a)^b &= (x + k cdot p)^b pmod p\ g^{ab} &= x^b + x^{b-1}(k cdot p) + cdots + (k cdot p)^b pmod p\ g^{ab} &= x^b pmod p end{align}

Similarly;

Let $g^b = y bmod p$ then $g^b = y + ell cdot p$ now take $a$-th power and $bmod p$ and arrive

$$g^{ab} = y^a pmod p$$

Therefore $$x^b = y^a = g^{ab} pmod p$$