I need to use Kolmogorov's forward equations for a pure birth process to derive the transition probabilities of the Yule process.
The equations for the pure birth process are \begin{align*} &P'_{ii}(t)=-\lambda_{i}P_{ii}(t)\\ &P'_{ij}(t)=\lambda_{j-1}P_{i, j-1}(t)-\lambda_{j}P_{ij}(t), \qquad j >i. \end{align*}
The problem is to show that $P_{ij}(t)={j-1 \choose i-1}e^{-\lambda i t}(1-e^{-\lambda t})^{j-i}$ for $j>i$. I have a hint to use induction on $j$.
So far, I've only used the fact that the Yule process has $\lambda_{i}=i\lambda$ to write the above equations as
\begin{align*} &P'_{ii}(t)=-i\lambda P_{ii}(t)\\ &P'_{ij}(t)=(j-1)\lambda P_{i, j-1}(t)-j\lambda P_{ij}(t), \qquad j >i. \end{align*} However, I'm not sure how to use the induction on $j$ hint. This is where I am stuck.
This formula is in fact valid for $j \ge i$. So we can start the induction at $j=i$: from the equation $P_{ii}'(t) = -i\lambda P_{ii}(t)$, as well as the obvious initial condition $P_{ii}(0)=1$, one has
$$P_{ii}(t) = e^{-i\lambda t} = {{i-1} \choose {i-1}}e^{-\lambda it}(1-e^{-\lambda t})^{i-i}.$$
Now assume $j>i$ and that the result is valid for $j-1$, that is,
$$P_{i,j-1}(t) = {{j-2}\choose{i-1}}e^{-\lambda it}(1-e^{-\lambda t})^{j-1-i}.$$
Then using your forward equations, we have the differential equation
\begin{align*} P_{ij}'(t) &= (j-1)\lambda{{j-2}\choose{i-1}}e^{-\lambda it}(1-e^{-\lambda t})^{j-1-i} -j\lambda P_{ij}(t), \\ P_{ij}(0) &= 0. \end{align*}
Now you just have to solve this differential equation. You can either rewrite the first line as $$ \frac{d}{dt}\Big(e^{\lambda jt}P_{ij}(t)\Big) = e^{\lambda jt}(j-1)\lambda{{j-2}\choose{i-1}}e^{-\lambda it}(1-e^{-\lambda t})^{j-1-i} = \lambda(j-i){{j-1}\choose {i-1}}e^{\lambda(j-i)t}(1-e^{-\lambda t})^{j-i-1},$$ then integrate, or simply verify that $$P_{ij}(t) = {{j-1}\choose{i-1}} e^{-\lambda it}(1-e^{-\lambda t})^{j-i}$$ solves this equation. I'll leave this to you.