Number of Permutations with given Cycle Types Let $n_1,\ldots ,n_k$ be the sizes of the 1st, 2nd,...,kth cycles in a permutation of $n$ elements. How is the number of permutations with this cycle structure and $n_1+\cdots + n_k=n$ equal to $${{n-1}\choose{n_1-1}}(n_1-1)!{{n-n_1-1}\choose{n_2-1}}(n_2-1)!\cdots{{n-n_1-\cdots-n_{k-2}-1}\choose{n_k-1}}(n_{k-1}-1)!$$

I understand that the chosen cycles can be permuted and that the elements of each cycle can be permuted, but I don't understand where all of the minus ones come from.
All of the $-1$'s come from the fact that the first element is chosen for you, in one way or another. For instance, when writing $\binom{n-1}{n_1-1}$, this is the number of ways to choose the elements in the first cycle. We have $-1$ everywhere because we already know that $1$ is part of that cycle. We are choosing the remaining $n_1-1$ elements fform the remaining $n-1$ in the set.

Likewise, $(n_1-1)!$ counts the number of different cycles you can make with those $n_1$ elements, because you put $1$ in front and then permute the remaining $n_1-1$ elements freely.

Next, with $n_2$, we choose the first element not chosen by $n_1$ (which can only be done one way) as well as $n_2-1$ elements freely from the remining $n-n_1-1$ elements. And so on.

Note that this way of counting sees $(123)(45)$ as having different cycle structure from $(12)(345)$, with the first having $n_1 = 3, n_2 = 2$ and the second having $n_1 = 2, n_2=3$.
All of the $-1$'s come from the fact that the first element is chosen for you, in one way or another. For instance, when writing $\binom{n-1}{n_1-1}$, this is the number of ways to choose the elements in the first cycle. We have $-1$ everywhere because we already know that $1$ is part of that cycle. We are choosing the remaining $n_1-1$ elements fform the remaining $n-1$ in the set.

Likewise, $(n_1-1)!$ counts the number of different cycles you can make with those $n_1$ elements, because you put $1$ in front and then permute the remaining $n_1-1$ elements freely.

Next, with $n_2$, we choose the first element not chosen by $n_1$ (which can only be done one way) as well as $n_2-1$ elements freely from the remining $n-n_1-1$ elements. And so on.

Note that this way of counting sees $(123)(45)$ as having different cycle structure from $(12)(345)$, with the first having $n_1 = 3, n_2 = 2$ and the second having $n_1 = 2, n_2=3$.