Question about $T_n = n + (n-1)(n-2)(n-3)(n-4)$ 
The formula $T_n = n + (n-1)(n-2)(n-3)(n-4)$ will produce an arithmetic sequence for $n < 5$ but not for $n \ge 5$. Explain why.

I think it is because if n is less than five the term with multiplication will be equal to zero and the common difference will be one. If n  is greater than five, the term added will have multiplication and there will not be a common difference.
Is this correct?

Your reasoning is correct. You can generalize it to say that
$$T_n = (an + b) + (n-1)(n-2)\cdots(n-k)$$
is an arithmetic progression for $1 \le n \le k$, for some arbitrary integer $k \ge 1$. This is because similarly, the product evaluates to $0$ for these values of $n$, to leave 
$$T_n = an + b$$
This remainder essentially defines an arithmetic progression with common difference $a$.
