Proving that the shortest simple path problem between two vertices $s$ and $t$ in a graph is NP-complete
In fact, it does not prove NP-completeness (see a list of our related reference questions).

For the shortest path problem (SPP)  to be NP-complete, it is crucial you allow negative edge weights. Then, there is a simple polynomial-time reduction from the Hamiltonian path problem to SPP. In other words, you take an arbitrary instance $I$ of the Hamiltonian path problem, and construct an instance $I'$ of SPP such that $I$ has a solution if and only if $I'$ has a solution. To make the reduction work, you only need to set up the edge weights in $I'$ suitably.