You always have the spontaneous emission rate. For stimulated emission it is multiplied by (n+1). This enhancement can be explained in terms of interference of the indistinguishable probability amplitudes leading to the same final result. If the emitted photon is in principle indistinguishable from the external field (that means you cannot tell afterwards which was the emitted photon and which photon was already present in the field) you have to sum up all probability amplitudes and square the result to get the emission probability. If the emission is distinguishable from the external field, you have to square all the probability amplitudes first and then sum them up. This is pretty similar to Feynman's explanation of the double slit. The enhancement in the indistinguishable boson case gives you the factor of (n+1) which is characteristic for stimulated emission. However, they can only be indistinguishable if they share the same quantum state. This requires them to have the same energy and also momentum.
edit: Maybe I should be a bit more clear. I assume the matrix element <n_{K'}+1|H|n_{K}> means that you start with an excited atom and a photon in state k and end up with a ground state atom, a photon in state k and a photon in state k' which has been in the vacuum state before. This is not forbidden. It is spontaneous emission.
Or to use a different line of reasoning: For Fermi's golden rule you need the density of final states. In the final state you have an atom in the ground state and two photons (let me call them a and b) leaving and you have an initial photon (called i) and an emitted photon (called e). So you can have the final states a=i,b=e or a=e,b=i. If the initial end emitted photon are distinguishable, these are indeed two separate final states, so you can apply Fermi's golden rule to calculate the transition probability for each of them. If they are indistinguishable, the states a=i,b=e and a=e,b=i are the same state. You have to add all indistinguishable processes in Fermi's golden rule before squaring them.
So you get either:
\frac{2 \pi}{\hbar}|\langle a=i,b=e|H|i\rangle^2 +\frac{2 \pi}{\hbar}\langle a=e,b=i|H|i\rangle|^2
or
\frac{2 \pi}{\hbar}|\langle a=i,b=e|H|i\rangle+\langle a=e,b=i|H|i\rangle|^2
Which makes a huge difference. This is somewhat like a bosonic exchange interaction or an interference effect. In my opinion it should not really be pictured as a photon hitting the atom and knocking out a photon or something similar (at least at an advanced level) as this can be misleading sometimes.
