There are a lot more than just those two; it's a fairly common exercise/lecture topic in an analysis course to prove that various definitions of completeness are equivalent, by building a directed graph of implications that allows us to reach anywhere from anywhere. For these two:
(1) implies (2): Take a sequence $x_n$ that is increasing and bounded above. By (1), it has a least upper bound $L$. We wish to show that $x_n$ converges to $L$. Since $L$ is an upper bound, it is $\ge$ every $x_n$. Since it's the least upper bound, $L-\epsilon$ is not an upper bound (for an arbitrary $\epsilon>0$) - which means that there is some $N$ such that $x_N > L-\epsilon$. Then, for all $n>N$, $L-\epsilon < x_N \le x_n\le L$ and $|x_n-L| <\epsilon$. That's the definition of convergence, and we have (2).
(2) implies (1): This one's harder - ideally, we'd take several steps around through other definitions of completeness. Still, we can do it in one step. For this, we use a successive bisection argument to find sequences converging to the least upper bound. Let $S$ be a nonempty set of real numbers bounded above by some $b$. Let $a$ be an element of $S$. Clearly, $a\le b$; if $a=b$, then it's the least upper bound and we're done. Assume otherwise.
Now, let $x_0=a,y_0=b$. Consider $\frac{x_0+y_0}{2}$. Is it an upper bound for $S$? If yes, let $x_1=x_0,y_1=\frac{x_0+y_0}{2}$. If no, let $x_1=\frac{x_0+y_0}{2},y_1=y_0$. Repeat this process; at each step, let $x_{n+1}=x_n, y_{n+1}=\frac{x_n+y_n}{2}$ if $\frac{x_n+y_n}{2}$ is an upper bound for $S$ and let $x_{n+1}=\frac{x_n+y_n}{2},y_{n+1}=y_n$ otherwise. The key property: for each $n$, $y_n$ is an upper bound for $S$ and $x_n$ isn't.
Now, the $x_n$ are an increasing sequence bounded above by $b$ and the $y_n$ are a decreasing sequence bounded below by $a$. They both have limits; let $x=\lim_n x_n$ and $y=\lim_n y_n$. In addition, $y_n-x_n=\frac{1}{2^n}(b-a)\to 0$.*
From those facts, the limits must be equal and $x=y$. We claim this common value is the least upper bound. It's an upper bound because the $y_n$ are; any element $c\in S$ is $\le$ each $y_n$, and thus also $\le$ their limit. On the other hand, any $z$ strictly less than $y$ is also strictly less than some $x_n$ - and since that $x_n$ isn't an upper bound, $z$ isn't either. That makes $y$ the least upper bound as desired. Done.
*Don't take this for granted. We quoted the archimedean property here; it's a standard lemma, but it does have to be proved if you're working from scratch.
