for all positive integers n. Yes, you can prove this using induction.
First, the base case n = 1:
1^1 \geq (1+1)^{1-1}
1 \geq 2^0
1 \geq 1
which is true.
Next, assume that the statement is true for some positive integer k, meaning
k^k \geq (k+1)^{k-1}
Now, we want to show that the statement is also true for k+1:
(k+1)^{k+1} \geq (k+2)^k
Expanding the left side, we get:
(k+1)^{k+1} = (k+1) \cdot (k+1)^k
Using our assumption, we can replace (k+1)^k with k^k \cdot (k+1), giving us:
(k+1)^{k+1} = (k+1) \cdot k^k \cdot (k+1)
Simplifying, we get:
(k+1)^{k+1} = k^{k+1} \cdot (k+1)^2
Now, we can use the fact that k \geq 1 to say that k+1 \geq 2. Therefore, (k+1)^2 \geq k+2, and we can replace it in our equation:
(k+1)^{k+1} \geq k^{k+1} \cdot (k+2)
Finally, using our assumption again, we can say that k^k \geq (k+1)^{k-1}, so:
(k+1)^{k+1} \geq (k+1)^{k-1} \cdot (k+2)
Which is true, since we are assuming k+1 \geq 2.
Therefore, by induction, the statement is true for all positive integers n. This means that n+1 \geq \left( 1+ \frac{1}{n} \right)^n is also true for all positive integers n, and you have proven your statement.
