Rotman's exercise 2.8 "$S_n$ cannot be imbedded in $A_{n+1}$"

I'm not sure what exactly I'm allowed to use or what Rotman had in mind, but I think this approach uses no more group theory than the (unique) factorization of permutations and the basic idea of what a homomorphism is.

If $n$ is odd there exists a subset $S\subset S_n$ of pairwise commuting elements of order $2$ with $|S|=\tfrac{n-1}{2}$. For example $$\{(1\ 2),(3\ 4),\ldots,(n-2\ n-1)\}.$$ An element $\sigma\in A_{n+1}$ of order $2$ is necessarily a product of an even number of disjoint $2$-cycles. Therefore a subset $A\subset A_{n+1}$ of pairwise commuting elements of order $2$ cannot contain more than $\tfrac{n+1}{4}$ elements. Hence for an embedding $S_n\ \hookrightarrow\ A_{n+1}$ to exist we must have $$\frac{n-1}{2}\leq\frac{n+1}{4},$$ or equivalently $n\leq3$. At this point it should not be hard to verify that $S_3$ does not embed into $A_4$, and of course that $S_1$ does embed into $A_2$.

A similar argument can be made when $n$ is even, to find that for an embedding $S_n\ \hookrightarrow\ A_{n+1}$ to exist we must have $$\frac{n}{2}<\frac{n+1}{4},$$ or equivalently $n<1$, immediately yielding a contradiction.