How to prove "For every positive integer n, $1^{n}$ + $2^{n}$ + ... + $n^{n}$ < $(n+1)^{n}$." by using induction? How can I prove the following theorem:

For every positive integer $n$, $1^n + 2^n + ... + n^n \lt (n+1)^n$

by using induction? 
I have proved that "for every real number $x > 0$ and every non-negative integer $n$, $x^{n} + n \cdot x^{n-1} \le (x+1)^n$".
It might be useful. Thank you.
 A: Let me try. Assuming that you have $$(P) \ \ \ \  \sum_{i=1}^k i^k < (k+1)^k$$. Now we prove that $$\sum_{i=1}^{k+1} i^{k+1} < (k+2)^{k+1}$$.
We have $$LHS = \sum_{i=1}^{k+1} i^{k+1} < (k+2)\sum_{i=1}^k i^k + (k+1)^{k+1} < (k+2)(k+1)^{k} + (k+1)^{k+1}< (k+2)^{k+1}.$$ (using your inequality with $x= k+1$)
A: Not your question, but you can do it directly:
\begin{align}(n+1)^n&=n^n+\binom{n}{n-1}n^{n-1}+\binom{n}{n-2}n^{n-2}+...+1\\
&\geq n^n+\binom{n}{n-1}(n-1)^{n-1}+\binom{n}{n-2}(n-2)^{n-2}+...+1 \\
&\geq n^n+(n-1)^{n-1}+ (n-2)^{n-2}+...+1.\end{align}
A: If $$1+2+・・・+(n-1)^n<n^n$$
when x=n+1 $$1+2+･･･+(n-1)^n+n^n<n^n+n^n<(n+1)^n$$
 that inequality works at x=n+1, too.  
A: This looks like the same proof as GAVD, but without sigma notation.  I didn't use your lemma explicitly either.

Let $P(n)$ be the statement that $1^n + 2^n + 3^n + \dots + n^n < (n+1)^n$.  Then $P(1)$ is $1 < 2$, which is definitely true.
Suppose $P(k)$ is true for some $k$.  That is, suppose
$$
    1^k + 2^k + \dots + k^k < (k+1)^k
$$
We want to show $P(k+1)$ is true.  Now
\begin{align*}
    1^{k+1} + 2^{k+1} + 3^{k+1} + \dots + k^{k+1} + (k+1)^{k+1} 
    &\leq(k+1)1^k + (k+1)2^k + \dots + (k+1)k^k \\&\quad\quad+ (k+1)(k+1)^k \\
    &=    (k+1)\left(1^k + 2^k + \dots + k^k + (k+1)^k\right) \\
    &\stackrel{(*)}{\leq}    (k+1)\left((k+1)^k + (k+1)^k\right)
     = 2(k+1)^{k+1}
\end{align*}
The point marked $(*)$ is where we used the inductive hypothesis.
Since $2 < k+1$, we have
$$
1^{k+1} + 2^{k+1} + 3^{k+1} + \dots + k^{k+1} + (k+1)^{k+1}
\leq (k+1)^{k+2}
$$
which establishes that $P(k+1)$ is true.  
Therefore, by induction, $P(n)$ is true for all $n$.
A: This inequality is obvious if you observe that
$$\frac1{(n+1)^{n+1}}\sum_{k=1}^n k^n=\frac1{n+1}\sum_{k=1}^n \Bigl(\frac k{n+1}\Bigr)^{\!n}$$
is the lower Riemann sum for the function $x^n$ on the interval $[0,1]$, with $n+2$ subdivision points. Thus
$$\frac1{n+1}\sum_{k=1}^n \Bigl(\frac k{n+1}\Bigr)^{\!n}<\int_0^n x^n\,\mathrm d\mkern1mu x=\frac1{n+1}\iff\sum_{k=1}^n \Bigl(\frac k{n+1}\Bigr)^{\!n}<1.$$
