I need a hint on a proof using mathematical induction I'm trying to prove that $k^k+1\ge2^k$ using mathematical induction but i'm missing something. How can i establish the binomial $(k+1)^{k+1}$? As a first step, i multiplied both sides by $2$ and $k$ but i can't get further than this without distorting things. Can someone give me a hint? Thanks in advance.
 A: Why bother use mathematical induction? For $k\ge2$, we have $k^k+1\ge2^k+1>2^k$.  

Maybe prove $k^k\ge2^k$ ($k\ge2$) would be much easier. Since $$(k+1)^{k+1}>(k+1)k^k\ge(k+1)2^k>2^{k+1},$$ mathematical induction can easily be applied.
A: $k=1$ $$\\ { 1 }^{ 1 }+1\ge 2\\$$ $k=n$ $$\\ { n }^{ n }+1\ge { 2 }^{ n }\\$$ ${ \left( n+1 \right)  }^{ n+1 }+1\ge \left( { n }^{ n+1 }+\left( n+1 \right) { n }^{ n }+...+1 \right) +1\ge { n }^{ n+1 }+1+{ n }^{ n }+1\ge { n }^{ n }+1+{ n }^{ n }+1=2\left( { n }^{ n }+1 \right) \ge { 2 }^{ n+1 }$
A: First for $k=1$ it is true, i.e., $1^1 + 1 \geq 2^1$. Suppose that the statement is true for $k$ want to show it is true for $k+1$. 
We have $k^k +1 \geq 2^k$ want to show that $(k+1)^{k+1} +1 \geq 2^{k+1}$ is true. 
$(k+1)^{k+1} = (k+1)^k \cdot (k+1)  $
Lets use the fact 
$(k+1)^k \geq k^k +1 $ to see this you can return to the binomial expansion. 
But we know that $k^k \geq 2^k +1 \Rightarrow k^k \geq 2^k-1$. Hence
$ k^k +1 \geq (2^k - 1) + 1\geq 2^k$. 
$(k+1)^{k+1} \geq 2^k (k+1) \geq 2^k (2)$. The last $\geq$ is true since $k\geq 1$.
Binomial Theorem https://en.wikipedia.org/wiki/Binomial_theorem
