I realize that there have been several answers to why $(t^a)$ is an ideal for the formal power series, but I was wondering why $(t+1)$ isn't an ideal? I'm rather new to the concept of ideals, so any help is appreciated!
$(t+1)$ means the ideal generated by $t+1$
$(t+1)$ is an ideal by your definition (that it is an ideal.)
In fact, $(t+1)=F[[t]]$ since $t+1$ is a unit.
The statement you are alluding to doesn't say that things of the form $(t^n)$ are the only ideals, or that they can only be expressed as $(t^n)$.
The zero ideal can't be expressed that way. And $(u)=(1)$ and $(ut^n)=(t^n)$ for any unit $u$ in $F[[t]]$.
Every formal power series can be written as
$$ f(t) = t^n \sum_{k = 0}^\infty a_k t^k $$
where $a_0 \ne 0$. But there is a theorem that tells us that if a power series has a non-zero constant term, then it has an inverse. For instance $(1 + t)^{-1} = 1 - t + t^2 - t^3 + t^4 - \cdots$ and if $A(t) = a_0 + B(t)t$ then
$$ \frac{1}{A(t)} = \frac{1}{a_0} \cdot \frac{1}{1 + a_0^{-1}B(t)t} = \frac{1}{a_0} \left( 1 - (a_0^{-1}B(t)t) + (a_0^{-1}B(t)t)^2 - (a_0^{-1}B(t)t)^3 +\cdots \right) $$
The factor of $t$ in $(a_0^{-1}B(t)t)^k$ ensures that this converges.
Thus if $f(t) = t^n A(t)$ where $A(t)$ has a nonzero constant term, then $(f) = (t^n)$ since $A(t)$ is a unit.
In general, you can show that if you have a non-zero ideal, $I$, then $I = (t^n)$ where $n$ is the smallest positive integer such that $I$ contains a series of the form $t^nA(t)$ with the constant term of $A(t)$ non-zero.
