A central series for a group $G$ is a sequence of normal subgroups
$$1 = G_1 \subset G_2 \subset \cdots \subset G_n = G$$
such that $G_{i+1}/G_i \subset Z(G/G_i)$. $G$ is nilpotent if it admits a central series. Recall that $G$ being solvable (i.e. admitting a normal series with abelian factors) is equivalent to a special normal series called the derived series $G \supset G' \supset G'' \supset \cdots$ eventually reaching 1. Likewise, $G$ being nilpotent is equivalent to any one of two special central series terminating. The first is the lower central series $G^1\supset G^2\supset \cdots$, defined by
$$G^1=G,\quad G^{i+1}=(G^i,G).$$
To prove it is indeed a central series we need to prove three things: $G^i\normal G$, $G^{i+1}\subset G^i$ and $G^i/(G^i,G)\subset Z(G/(G^i,G))$. The first follows by induction: supposing $G^i\normal G$ we have
$$g(G^i,G)g\inv \subset (gG^ig\inv, gGg\inv) \subset (G^i,G)\quad \text{for all $g\in G$}.$$
The second fact follows from the first, because given $h\in G^i$ and $g\in G$ we have $gh\inv g\inv\in G^i$ and so
$$(h,g) = h(gh\inv g\inv)\in G^i.$$
The last follows from a more general observation which characterises subgroups of the form $(H,G)$. In fact, it generalises the characterisation of $G'$ as the smallest normal subgroup yielding an abelian quotient.
Proposition 1. Let $H,N\normal G$, and $H/N$ denote the image of $H$ under the projection $G\to G/N$. (Think of $G\Normal H$ as fixed and $N$ as varying.) Then
$$H/N\subset Z(G/N) \iff (H,G)\subset N.$$
Proof. Note that $(H,G)\normal G$ has been shown above, in the context $H=G^i$. The proof is just definition-chasing:
$$\begin{align*} H/N \subset Z(G/N) &\iff (hN)(gN)=(gN)(hN)\ \text{for all $h\in H, g\in G$} \\ &\iff (hN,gN) = 1 \\ &\iff (h,g)\in N \\ &\iff (H,G)\subset N.\ \qed \end{align*}$$
This result justifies the descriptor 'lower', since by the proposition $(G^i,G)$ is the smallest subgroup $N$ satisfying the property $G/N\subset Z(G/N)$. Next we have the upper central series $C_1\subset C_2\subset\cdots$, defined upwards by
$$C_1=1,\quad C_{i+1}\ \text{is the unique subgroup $\supset C_i$ satisfying $C_{i+1}/C_i=Z(G/C_i)$}.$$
In particular, $C_2=Z(G)$. It is quite easy to see why this series is given the descriptor 'upper'. The conditions $C_i\subset C_{i+1}$, $C_{i+1}/C_i\subset Z(G/C_i)$ are satisfied by definition, and we have $C_{i+1}\normal G$ since $C_{i+1}$ is the kernel of a certain map $G\to (G/C_i)/(C_{i+1}/C_i)$.
I now prove that $G$ being nilpotent is equivalent to either the lower central series terminating at 1, or the upper central series terminating at $G$.
Proposition 2. $G$ nilpotent $\iff$ lower central series terminates.
Proof. Backward direction is trivial. As for the forward direction, let $G=G_1\supset G_2\supset \cdots \supset G_n=1$ be a central series. I'll show by induction that $G^i\subset G_i$ for all $i$. Suppose this is the case for some $i$. Then $G\supset G_i\supset G_{i+1}$ is a tower of normal subgroups satisfying $G_i/G_{i+1}\subset Z(G/G_{i+1})$, and so by Proposition 1 we have $G_{i+1}\supset(G_i,G)$, and by the induction hypothesis we can conclude $G_{i+1}\supset (G_i,G)\supset (G^i,G) = G^{i+1}$. $\qed$
Proposition 3. Lower central series terminates $\iff$ upper central series terminates.
Proof. Backward direction follows from Proposition 2. As for the forward direction, suppose $1=G^n\subset G^{n-1}\subset \cdots\subset G^1=G$ is the lower central series and $1=C_1\subset C_2\subset\cdots$ is the upper central series. I will show by induction that $C_i\supset G^{n-i}$. Suppose this is the case for $i$. Then consider the normal subgroups $G^{n-i-1},C_i\normal G$. Since $C_i\supset G^{n-i} = (G^{n-i-1},G)$, by Proposition 1 we have $G_{n-i-1}/C_i\subset Z(G/C_i) = C_{i+1}/C_i$ and so $G_{n-i-1}\subset \pi^{-1}(G^{n-i-1}/C_i)\subset C_{i+1}$, where $\pi$ is the projection onto $C_i$. $\qed$
Finally, I outline the relationship between nilpotent and solvable. All nilpotent groups are solvable, because the lower central series terminates and $G^{(i)}\subset G^i$ for all $i$, so the derived series also terminates. However, not all solvable groups are nilpotent. In fact, any solvable group with trivial center will do, because the upper central series will never terminate. The simplest example is $S_3$. The center is the set of elements fixed by conjugation by every element, but by the behaviour of conjugation in symmetric groups, it shouldn't be hard to see that no such element exists. However, $S_3$ admits the normal series $S_3\Normal A_3\Normal 1$ with abelian factors.