By calculating the first few terms of N, I think the sum
$$\sum_{k=0}^N \binom{N}{k} \prod_{j=1}^k [ (j-1)\lambda + (N-1)\mu ] \prod_{i=1}^{N-k} [(i-1)\lambda + (N-1)\gamma ]$$
is equal to
$$ \prod_j^N [(j-1) \lambda + (N-1)\mu + (N-1) \gamma] $$
But I'm not sure about how to prove this is true.
We can use Pochhammer notation to simplify the problem.
Define $(x)_n=x(x+1)...(x+n-1)$
$$\frac{N!}{k!(N-k)!} \lambda ^k \lambda ^{N-k} \left(\frac{(N-1)\mu}{\lambda}\right)_k \left(\frac{(N-1) \gamma}{\lambda}\right)_{N-k}$$
Then using the relation $(x)_n=n! \binom{x}{n}$
$$\lambda^N N! \sum_{k=0}^N \binom{\frac{(N-1)\mu}{\lambda}}{k} \binom{\frac{(N-1)\gamma}{\lambda}}{N-k}$$
We can then apply the Chu–Vandermonde identity $\sum_{k=0}^N \binom{m}{k} \binom{n-m}{N-k}=\binom{n}{N}$
$$\lambda^N N! \binom{\frac{(N-1)(\mu+\gamma)}{\lambda}}{N}$$ $$\lambda^N \prod_{j=1}^N \left( \frac{(N-1)(\mu+\gamma)}{\lambda} - j +1 \right)$$
$$\prod_{j=1}^N \left((j-1)\lambda + (N-1)(\mu+\gamma) \right)$$
Which is the desired expression.
