$(1+p)^n$ is not $1 \pmod {p^r}$ when $n < p^{r-1}$ Let $p$ be an odd prime. I know that $(1+p)^{p^{r-1}}\equiv 1 \pmod {p^r}$ but how can I prove that if $n < p^{r-1}$ then $(1+p)^n$ is not $1 \pmod {p^r}$. I tried to prove using the properties of binomial expansion but I cannot show either the sum is less than $p^r$ or not $1 \bmod p^r$. 

Since you know $(1+p)^{p^{r-1}}\equiv 1\pmod{p^r}$, you know that the order of $1+p$ mod $p^r$ divides $p^{r-1}$. So for your problem it suffices to check that $(1+p)^{p^e}\not\equiv1\pmod{p^r}$ for any $e<r-1$, or better, just that 
$$(1+p)^{p^{r-2}}\not\equiv1\pmod{p^r}.$$
But now the binomial theorem gives 
$$(1+p)^{p^{r-2}}=\sum_{k=0}^{p^{r-2}}\binom{p^{r-2}}{k}p^k=1+\sum_{k=1}^{p^{r-2}}\binom{p^{r-2}}{k}p^k\equiv
1+p^{r-1}\not\equiv 1\mod p^r.$$
