In Cameron's Combinatorics, he set out to proof the below identity:
$$\sum_{n\ge0}\frac{S(n,k)t^n}{n!}=\frac{(exp(t)-1)^k}{k!}$$
He proceeds as follows:
$$\begin{align*}\sum_{n\ge0}\frac{S(n,k)t^n}{n!}&=\sum_{n\ge0}\frac{1}{k!}\sum_{j=1}^k{k\choose j}(-1)^{k-j}j^n\frac{t^n}{n!}\\ &=\frac{1}{k!}\sum_{j=1}^k{k\choose j}(-1)^{k-j}\sum_{n\ge0}\frac{{(jt)}^n}{n!}\\ &=\frac{1}{k!}\sum_{j=1}^k{k\choose j}(-1)^{k-j}exp(jt)\\ &=\frac{(exp(t)-1)^k}{k!}\end{align*}$$
I had no trouble following author's proof until the second-to-last line. I believed the author used the binomial theorem to prove the equivalence of that expression to the last one. However, the starting index of the summation in the binomial theorem is $0$ instead of $1$: $$(1+t)^n=\sum_{k=0}^n{n \choose k}t^k$$
That is, if I tried to convert the numerator of the aforementioned expression using binomial theorem: $$\begin{align*}(exp(t)-1)^k&=(-1)^k(1-exp(t))^k\\ &=(-1)^k(1+(-exp(t)))^k\\ &=(-1)^k\sum_{j=0}^k{{n\choose j}(-1)^{-j}(exp(t))^j}\\ &=\sum_{j=0}^k{k\choose j}(-1)^{k-j}exp(jt) \end{align*}$$
I think I made a mistake somewhere but I don't know where. I am completed stuck here in trying to understand the proof. Any help is greatly appreciated. Thank you in advance.
You did not make any mistakes, but the author did. The author's mistake was when they wrote $$ \sum_{n\ge0}\frac{S(n,k)t^n}{n!} \stackrel{?}=\sum_{n\ge0}\frac{1}{k!}\sum_{\color{red}{j=1}}^k{k\choose j}(-1)^{k-j}j^n\frac{t^n}{n!}\tag1 $$ They are, presumably, basing this off of the formula $$ S(n,k)\stackrel?=\frac1{k!}\sum_{\color{red}{j=1}}^k \binom kj(-1)^{k-j}j^n,\tag2 $$ which is incorrect. The lower index for the summation here is supposed to be $j=0$. That is, the correct formula is $$ S(n,k)=\frac1{k!}\sum_{\color{green}{j=0}}^k \binom kj(-1)^{k-j}j^n\tag3 $$ When you make that replacement, everything works out just fine. Note that $(2)$ and $(3)$ are almost exactly equal to each other; most of the time, $j^n=0$ when $j=0$, so it is fine to omit the $j=0$ term. However, since $0^0=1$ (in the context of combinatorics), you cannot omit this term when $n=0$. This makes a difference in $(1)$, because we are summing over all $n\ge 0$.