About the special linear group $SL(n,\,\mathbb{Z})$ I found the following relation: $$SL_n(\mathbb{Z})=\langle e+e_{12},\, e_{12}+\dots+e_{n-1, n}+(-1)^{n-1}e_{n1}\rangle$$
where $e_{ij}$ is the matrix with its $(i,\, j)$th entry $1$, and all other entries $0$; and $e$ is the identity.
How one can prove that? Can you give me any clue about it?
References


*

*This is the problem 2.5 of Kargapolov's Fundamentals of Group Theory

*It is stated here on page $5$.
 A: Notice how these generators modify a given matrix when you multiply that matrix from the right: The first adds row 2 to row 1 (and its inverse subtracts row 2 from row 1). And the other rotates the rows (while flipping one sign). 
Let $S$ be the set of pairs $(i,j)$ such that it is possible to add row $j$ to row $i$ (and by existence of inverses one can also subtract row $j$ from row $i$). The first generator gives us immediately that $(1,2)\in S$. Conjugation with the second generator shows that $(i,j)\in S$ implies $(i+1,j+1)\in S$, so by now $(i,i+1)\in S$ for al $i$ (with $(n,n+1)$ understood to mean $(n,1)$). Assume that $d<n-1$ and we have $(i,j)\in S$ for all $i,j$ with $j=i+d$. Then also $(1,d+2)\in S$: add row $d+2$ to row $2$; then add row $2$ to row $1$; then subtract row $d+2$ from row $2$; then add row $2$ to row $1$. Combinig this we find that $(i,j)\in S$ for all $1\le i<j\le n$. Via conjugation with the rotation generator we obtain that $(i,j)\in S$ whenever $i\ne j$.
This allows you to swap any two rows (at the expense of introducing a sign: $(a,b)\to (a+b,b)\to (a+b,-a)\to (b,-a)$.
Now we have enough tools at hand to run an almost-Gauss algorithm: As long as at least  two rows  have nonzero first column entries, add/subtract one to/from the maximal (in absolute value) one such that $\sum |a_{i,1}|$ decreases. After finitely many steps, decreasing is no longer possible, which means that all but one row has $a_{i,1}=0$ now and the remaining row has a $\pm1$ entry. Swap the nonzeor entry to row $1$ and proceed with simplyfying column 2, then 3, etc. This takes you to an upper triangle matrix, from which you readily obtain a diagonal matrix with all diagonal entries $\in\{1,-1\}$. Since the determinant is positive, the number of $-1$ entries must be even. Swapping two such rows twice flips both signs, thus reducing the number of negative entries. Ultimately we obtain the identity matrix. 
This was to be shown (wasn't it?)
