I'm having some trouble understanding one part of the derivation of path integral formulation of QM, say I have the propagator, then we can break it up in N parts s.t.
$$[x',t_1;x_0,t_0]=[x';e^{i\textbf{H}\Delta t/\hbar}...._{N\ times}....e^{i\textbf{H}\Delta t/\hbar};x_0], \tag{1}$$
defining $[a;b]$ to be bra-kets, respectively and $\textbf{H}$ our hamiltonian operator, where we're taking particle from point $x_0$ at $t_0$ to point $x'$ at $t_1$ and the intervals are broken giving us $\Delta t=(t_1-t_0)/N$.
Thus, we can rewrite as
$$[x',t_1;x_0,t_0]=[x_N;e^{i\textbf{H}\Delta t/\hbar};x_{N-1}]...[x_1;e^{i\textbf{H}\Delta t/\hbar};x_{0}] \tag{2}$$ For me, we'd have $$[x_N;e^{i\textbf{H}\Delta t/\hbar};x_{N-1}]=\int dx_{N-1}[x_N;e^{i\textbf{H}\Delta t/\hbar};x_{N-1}][x_{N-1};x_{N-1}]; \tag{3}$$ Where I used the identity $$1 = \int dx\ ;x_N][x_N; \tag{4}$$
I think we would have something like this equation after solving (2) in integral form:
$$[x',t_1;x_0,t_0]= $$ $$ \int dx_{N-1}[x_N;e^{i\textbf{H}\Delta t/\hbar};x_{N-1}][x_{N-1};x_{N-1}] \int dx_{N-2}[x_{N-1};e^{i\textbf{H}\Delta t/\hbar};x_{N-2}][x_{N-2};x_{N-2}]... \\ \times\int dx_{0}[x_1;e^{i\textbf{H}\Delta t/\hbar};x_{0}][x_{0};x_{0}] \qquad (5) $$
but books are saying the following:
$$[x',t_1;x_0,t_0]=$$ $$ \int dx_1 ... \int dx_{N-1}[x_N;e^{i\textbf{H}\Delta t/\hbar};x_{N-1}][x_{N-1};e^{i\textbf{H}\Delta t/\hbar};x_{N-2}]...[x_2;e^{i\textbf{H}\Delta t/\hbar};x_{1}] [x_1;e^{i\textbf{H}\Delta t/\hbar};x_{0}] $$
And I quite can't see why.
So here are my questions:
Did I do any mistake in my derivation?
Why last equation holds? Instead of having the propagators inside of each integral.