# Obtaining $\beta = \frac{1}{k_B T}$ from first principles derivation

I shall try to explain my understanding of how this problem should be solved. First, I agree with your derivation of
$$n_i = N e^{\beta E_i}/Z, \qquad (1)$$ where $$Z = \sum_{j=1}^r e^{\beta E_j} \qquad (2)$$ and $$\beta$$ is obtained from equation $$\frac{E}{N} = \frac1Z \sum_{i=1}^r E_i e^{\beta E_i}. \qquad (3)$$ We can make a conclusion now, that $$n_i$$ and $$\beta$$ depend on $$E$$ and $$N$$.

Our goal is to show that we have Gibbs distribution for $$n_i$$ and $$\beta$$ is equal to $$-1/k_BT$$. How can we relate $$\beta$$ and $$T$$? The only way I know is to use the following thermodynamic relation: $$\frac1T = \left(\frac{\partial S}{\partial E}\right)_N, \qquad (4)$$ To obtain $$S$$ as a function of $$E$$ and $$N$$, we shall use Boltzmann's formula $$S = k_B\log\Omega.$$ Then $$S = S_0(N) -k_B\sum_{i=1}^r n_i\log(n_i/e). \qquad (5)$$ here $$n_i$$ depend on $$E$$ and $$N$$. Variation of $$S$$ induced by variations of $$E$$ and $$N$$ can be expressed through variations of $$n_i$$: $$\delta S = S_0'(N)\sum_{i=1}^r \delta n_i - k_B\sum_{i=1}^r\delta n_i \log(n_i). \qquad (6)$$ Here $$\delta n_i$$ are expressed in terms of $$\delta E$$ and $$\delta N$$ in a somewhat cumbersome way. Substitution of (1) into (6) gives $$\delta S = -k_B\beta\sum_{i=1}^r E_i\delta n_i + \mbox{"smthng"}\sum_{i=1}^r \delta n_i. \quad (8)$$ Obviously, $$\delta n_i$$ satisfy equations: $$\sum_{i=1}^r \delta n_i = \delta N, \qquad \sum_{i=1} E_i \delta n_i = \delta E.\quad (9)$$ Then (8) and (9) give $$\delta S = -k_B\beta\delta E + \mbox{"smthng"}\delta N. \quad (10)$$ The last equation gives $$\left(\frac{\partial S}{\partial E}\right)_N = -k_B\beta. \quad (11)$$ At last, (11) and (4) give: $$\beta = -\frac1{k_BT}.$$

Update. My explanation just demonstrates the statement of @higgsss's comment.