Why $1\in I\implies I=R$?

Question

Written in Abstract Algebra by T. W. Judson :

Theorem

Let R be a ring with identity and suppose that I is an ideal in R such that 1 is in I. Since for any r∈R, r1=r∈I by the definition of an ideal, I=R.

and considering the definition of the ideal of a ring in the same book:

Definition

An ideal in a ring R is a subring I of R such that if a is in I and r is in R, then both ar and ra are in I; that is, rI⊂I and Ir⊂I for all r∈R.

My question is : How $1$ being an element of $I$ implies that $I$ is $R$ itself ?!

Thanks a lot.

PS It is not written as a theorem title, the claim was in the middle of the text with no mentioning of a proof.

Answer 1

Any $R$-multiple of an element of $I$ is again in $I$ (by definition). So let $r\in R$, then $r\cdot 1$ as a product of an element $r\in R$ and $1\in I$ is again in $I$, hence $I=R$.

Answer 2

Because for any $r\in R$ we have $r=r1\in I\subseteq R$.