Prove that the right shift operator $S(x_1,x_2, \dots) = (0,x_1,x_2,\dots)$ is bounded. Exercise :

Let $S: l^1 \to l^1$ be the right-shift operator :
  $$S(x_1,x_2,\dots) = (0,x_1,x_2,\dots)$$
  Prove that $S$ is bounded and find its norm.

Attempt :
The space $l^1$ is : $$l^1 = \{(x=(x_n) : \sum_{i=1}^\infty x_n < + \infty\}$$
To show that the operator $S$ is bounded, I must show that :
$$\exists \; M>0 : \|Sx\|\leq M\|x\|$$
But I can't really see how to proceed on this particular proof without having any knowledge of the norm that should be used. Shall the norm of $l^1$ space be used ? If so, what does  "find the norm of the operator S" ? 
I would really appreciate any tips/solutions or clarifications regarding this particular exercise.
Note : I have NOT been introduced to isometries in my Functional Analysis class yet, so I am looking for an elementary bounded operator approach.

$$\|S(x_1,x_2,\dots)\|_1=\|(0,x_1,x_2,\cdots)\|_1=0+|x_1|+|x_2|+\dots = |x_1|+|x_2|+\dots = \|(x_1,x_2,\dots)\|_1.$$
In particular, $$\|S(x_1,x_2,\cdots)\|_1\le 1\cdot \|(x_1,x_2,\cdots)\|_1.$$
This inequality says that $S$ is bounded and $\|S\|\le 1$.
But $S(1,0,0,\dots)=(0,1,0,\dots)$, so $$\|S(1,0,\dots)\|_1 =\|(0,1,0,\dots)\|_1=1=\|(1,0,\dots)\|_1.$$
So $\|S\|=1$
