Integer which is Multiplied by 9 when moving Last Digit to First/Proof 1

Proof
First we demonstrate that $N$ has this property:

A way to find $N$ is given below.

Suppose $N = \sqbrk {a_k a_{k - 1} \dots a_1}$ satisfies the property above.

Consider the rational number represented by $q = 0.\dot {a_k} a_{k - 1} \dots \dot {a_1}$.

Since $a_k \ne 0$, we require $q \ge 0.1$.

By the property we have:
 * $9 q = 0 . \dot {a_1} a_k \dots \dot {a_2}$

Then:
 * $90 q = a_1 . \dot {a_k} a_{k - 1} \dots \dot {a_1}$

Notice that the recurring part of $90 q$ is the same as the recurring part of $q$.

Subtracting, we get $89 q = a_1$.

We have:
 * $a_1 \ge 89 \times 0.1 = 8.9$

so:
 * $a_1 = 9$

which gives:
 * $q = \dfrac 9 {89}$

Decimal Expansion
which gives us:
 * $N = 10 \, 112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 808 \, 988 \, 764 \, 044 \, 943 \, 820 \, 224 \, 719$

as the smallest positive integer satisfying the property.