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.

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 $n$.
