Eulerian graph theorem How can I prove the following theorem:   
For a connected multi-graph
G, G is Eulerian if and only if every vertex has even degree.  

I found a proof here: in this PDF file, but, it merely consists of language that is very hard to follow and doesn't even give a conclusion that the theorem is proved. So, how can I prove this theorem?
 A: Def: An Eulerian cycle in a finite graph is a path which starts and ends at the same vertex and uses each edge exactly once. 
Def: A finite Eulerian graph is a graph with finite vertices in which an Eulerian cycle exists.
Def: A graph is connected if for every pair of vertices there is a path connecting them. 
Def: Degree of a vertex is the number of edges incident to it. 
Claim: A finite connected graph is Eulerian iff all of its vertices are even degreed. 
Pf: 
($\Longrightarrow$)
Let $G=(V,E)$ be a connected Eulerian graph. By def. of being an Eulerian graph, there is an Eulerian cycle $Z$, starting and ending, say, at $u\in V$. Each visit of $Z$ to an intermediate vertex $v\in V\setminus\{u\}$ contributes 2 to the degree of $v$, so each $v\in V\setminus\{u\}$ has an even degree. As for $u$, each intermediate visit of $Z$ to $u$ contributes an even number, say $2k$ to its degree, and lastly, the initial and final edges of $Z$ contribute 1 each to the degree of $u$, making a total of $1+2k+1=2+2k=2(1+k)$ edges incident to it, which is an even number. 
($\Longleftarrow$) (By Strong Induction on $|E|$)
B.S.: $|E|=0$. Since $G$ is connected, there must be only one vertex, which constitutes an Eulerian cycle of length zero.
I.H.: The claim holds for all graphs with $|E|<n$. 
I.S.: Let $G$ be a graph with $|E|=n\in \mathbb{N}$. 
Def: A tree is a graph which does not contain any cycles in it. 
Def: A spanning tree of a graph $G$ is a subset tree of G, which covers all vertices of $G$ with minimum possible number of edges.
Def: A leaf is a vertex of degree 1.
Since $G$ is connected, there should be spanning tree $T=(V',E')$ of $G$. 
Lemma: A tree on finite vertices has a leaf. Pf: Let $V=\{v_1,\ldots, v_n\}$. For a contradiction, let $deg(v)>1$ for each $v\in V$. Now start at a vertex, say $v_{i_1}$. Now 'walk' over one of the edges connected to $v_{i_1}$ to a vertex $v_{i_2}$. Since the degree of $v_{i_2}$ is 2, we can walk to a vertex $v_{i_3}\neq v_{i_2}$ and continue this process. Since $V$ is finite, at a given point, say $N$, we will have to connect $v_{i_N}$ to $v_{i_1}$, and have a cycle, $(v_{i_1}, \ldots, v_{i_N}, v_{i_1})$, contradicting the hypothesis that $G$ is a tree. 
Hence our spanning tree $T$ has a leaf, $u\in T$. (It might help to start drawing figures from here onward.) Since $deg(u)$ is even, it has an incidental edge $e\in E\setminus E'$. Now consider the cycle, $C:=(V',E\cup\{u\})$. Let $G':=(V,E\setminus (E'\cup\{u\}))$. 
Clearly, $deg_{G'}(v)= \left\{\begin{array}{lr}
        deg_G(v)-2, & \text{if } v\in C\\
        deg_G(v), & \text{if } v\notin C
        \end{array}\right.$ 
Suppose $G'$ consists of components $G_1,\ldots, G_k$ for $k\geq 1$. By Inductive Hypothesis, each component $G_i$ has an Eulerian cycle, $S_i$. Also each $G_i$ has at least one vertex in common with $C$. Let $x_i\in V(G_i)\cap V(C)$. By a renaming argument, we may assume that $S_i$ begins with $x_i$ and ends at $x_i$, since $S_i$ passes all edges in $G_i$ in a cyclic manner. Now, a traversal of $C$, interrupted at each $x_i$ to traverse $S_i$ gives an Eulerian cycle of $G$.
