Applying a bra on the density operator

Question

the density operator can be written as $$ \rho=\sum^N _{i=1} w_i |i\rangle\langle i| $$ Now I am not sure if the following is true $$ \langle k|\rho|k\rangle=\langle k|\bigg(\sum^N _{i=1} w_i |i\rangle\langle i|\bigg)|k\rangle=\sum^N _{i=1} w_i |\langle i|k \rangle|^2 $$ I know that $ |i \rangle \langle i| |k\rangle = \langle i|k \rangle |i\rangle$ so you can bring the right $|k\rangle$ into the sum and connect it to $\langle i|$
But I have never seen the rule $\langle k| |i\rangle\langle i|=\langle i|\langle k|i \rangle$. It should be true I guess. But I cant come up with a way to proof it

Answer 1

$\langle k|i\rangle$ is just a complex number, and can be moved around freely. In the same way that $3\langle i| = \langle i |3 $, you have that $\langle k|i\rangle\langle = |i\rangle \langle k|i\rangle$.

Answer 2

I'm not sure much is going through anything except for the coefficient $w_i$, which is a scalar and can be moved freely (keeping indexed sums coherent).
It's a straightforward example of norm of states $$ \langle k|\rho|k\rangle=\langle k|\bigg(\sum^N _{i=1} w_i |i\rangle\langle i|\bigg)|k\rangle=\sum^N _{i=1} w_i \langle k|i \rangle\langle i|k \rangle=\\=\sum^N _{i=1} w_i \langle i|k \rangle^\dagger\langle i|k \rangle=\sum^N _{i=1} w_i |\langle i|k \rangle|^2 $$ where the square norm i have defined is through Hermitian conjugation. This is usually the case, but it can also be defined via integral relations, complex conjugation etc.