A memoryless property means:
$$P(X \gt t + s | X \gt s) = P(X \gt t)$$
So we can use this to state:
$$\begin{align} P(X \gt t + s) &= P(X \gt s) P(X \gt t + s | X \gt s) \\ &= P(X \gt s) P(X \gt t) \end{align}$$
Let \(G_x(t) = P(X \gt t)\):
$$G(t + s) = G(s) G(t)$$
If \(s = t\)
$$G(2t) = G(t)^2$$
If \(s = 2t\)
$$G(3t) = G(2t)G(t) = G(t)^3$$
In general \(G(kt) = G(t)^k\). Now let's say we raise both sides to the power of \(1/k\), and replace \(t\) with \(h/k\):
$$G(h)^{\frac{1}{k}} = G \left(\frac{h}{k} \right)$$
This means:
$$G(j)^{\frac{m}{k}} = G \left(j\frac{m}{k} \right)$$$$G(j)^x = G \left(jx \right) \text{ for all rational } x \gt 0$$
Let \(j=1\):
$$G(x) = G(1)^x$$$$P(X \gt x) = e^{xln(G(1))}$$
Since \(G(1)\) is between 0 and 1, then \(ln(G(1))\) is some negative constant:
$$P(X \gt x) = e^{-\lambda x}$$
I got my derivation from this video. From the above, we can get the cumulative distribution function:
$$F(x) = P(X \le x) = 1 - e^{-\lambda x}$$
We can derivate this to get the pdf:
$$f(x) = \lambda e^{-\lambda x}$$