Let us recall that the Stirling numbers satisfy the identities: $$\begin{align} \displaystyle{ n \brace k} &= \frac{1}{k!}\sum_{j=0}^k (-1)^{k-j}{k \choose j} j^n = \frac{1}{k!} \sum_{j=0}^k (-1)^j {k \choose j}(k-j)^n \\[6pt] { n+1 \brace k} &= k{ n \brace k} + { n \brace k-1} \end{align}$$
and appear in the Taylor expansion: $$\frac{(e^w-1)^k}{k!} = \sum_{n=k}^\infty { n \brace k}\frac{w^n}{n!}.$$
i am confused about the last equation. through researching, ive done converting $(e^w-1)^k$ into a binomial expansion then i got $∑(−1)^{k−j} \binom k j e^{jw}$ and convert $e^{jw}$ then i dont know how to multiply or do some mathematical operations with sigma notation, i dont know what is next..how can i convert that into $$\frac{(e^w-1)^k}{k!} = \sum_{n=k}^\infty { n \brace k}\frac{w^n}{n!}.$$
\begin{align} \frac{(e^{w}-1)^{k}}{k!}&=\frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}e^{jw}=\frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}\sum_{m=0}^{\infty}\frac{(jw)^{m}}{m!} \\ &= \frac{1}{k!}\sum_{j=0}^{k}\sum_{m=0}^{\infty}(-1)^{k-j}\binom{k}{j}\frac{(jw)^{m}}{m!}=\sum_{m=0}^{\infty}\left(\frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}j^{m}\right)\frac{w^{m}}{m!}\\ &=\sum_{m=0}^{\infty} { m \brace k}\frac{w^{m}}{m!}=\sum_{m=k}^{\infty} { m \brace k}\frac{w^{m}}{m!} \end{align}